高速路网-爱来自第三学期

这是一个项目题。

题目描述

2024-07-12T07:28:47.png 2024-07-12T07:28:57.png 2024-07-12T07:29:09.png 2024-07-12T07:29:18.png

题意分析

分为2个部分: 1.将图片转换为Graph 2.对Graph进行次短路算法

实现思路:

1.将图片转换为Graph

1.对读取的PNG进行广度优先染色,相同颜色的标一个id,并将每一个州的中心点记录 2.对染色后的地图的每一个州进行六个方向的查询,进行建边

2.次短路

使用迷阵题的代码,注意如果有多条最小路径,不能输出最小路径,要输出次小路径。 使用优先队列优化时间。

代码:

小学期结束后填坑。

https://se.jisuanke.com/_996/graph_lab(已填坑)

#include "state.h"
#include <queue>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <set>
void init_State(struct State *s)
{
    s->industry = nullptr;
    s->pre = nullptr;
    s->count = 0;
    s->cord.push_back({0, 0});
    s->mind = 0;
    s->Edge.clear();
    // TODO
}
void delete_State(struct State *s)
{
    // TODO
    delete[] s->industry;
    delete[] s->pre;
    std::vector<std::vector<edge>>().swap(s->Edge);
    std::vector<std::pair<int, int>>().swap(s->cord);
}
void parse(struct State *s, struct PNG *p)
{
    // TODO
    // 即统计图片中联通块的个数,或者叫做统计州
    // s中要存所有州的信息,以及他们的相邻关系
    // 便于后续查找最短路径
    // p->image 为读入的位图数据
    // 思路:循环对每个结点进行编号
    // 对每个结点进行6个方向的循环加边
    // p->image max_size 5000*5000

    // -1代表白色 -2代表黑色 1-n代表该像素块属于的编号
    int *ids = new int[p->width * p->height]();

    int id = 1;
    for (int y = 0; y < p->height; y++)
    {
        for (int x = 0; x < p->width; x++)
        {
            if (ids[y * p->width + x] == 0)
            {
                PXL *pixel = get_PXL(p, x, y);
                if (pixel->blue == 0 && pixel->green == 0 && pixel->red == 0)
                { // black
                    int tmpx, tmpy;
                    color(p, ids, -1, x, y, tmpx, tmpy);
                }
                else if ((pixel->red == 255 && pixel->blue == 255 && pixel->green == 255))
                { // white
                    int tmpx, tmpy;
                    color(p, ids, -2, x, y, tmpx, tmpy);
                }
                else
                { // 正常格子
                    int tmpx, tmpy;
                    color(p, ids, id, x, y, tmpx, tmpy);
                    s->cord.push_back({tmpx, tmpy});
                    id++;
                }
            }
        }
    }
    // 染色标记完毕

    // 开始记录每个州的发达程度industry
    s->count = id - 1;
    s->industry = new int[s->count + 1]();
    s->Edge.resize(s->count + 1);
    for (int y = 0; y < p->height; y++)
    {
        for (int x = 0; x < p->width; x++)
        {
            if (ids[y * p->width + x] != -1 && ids[y * p->width + x] != -2)
            {
                int _id = get_id(ids, x, y, p->width, p->height);
                if (s->industry[_id] != 0)
                {
                    continue;
                }
                PXL *pixel = get_PXL(p, x, y);
                s->industry[_id] = cal_industry(pixel);
            }
        }
    }

    // 对每个州进行六方向连接 建边  标记某个顶点是否已经扩展好边
    int dir[6][2] = {
        {-8, 0}, {8, 0}, {-4, -8}, {4, -8}, {-4, 8}, {4, 8}};
    for (size_t i = 1; i <= s->count; i++)
    {
        std::pair<int, int> cord = s->cord[i];
        for (int j = 0; j < 6; j++)
        {
            int tx = cord.first + dir[j][0];
            int ty = cord.second + dir[j][1];

            if (in(tx, ty, p->width, p->height))
            {
                int toid = get_id(ids, tx, ty, p->width, p->height);
                if (toid != 0 && toid != -1 && toid != -2)
                {
                    s->Edge[i].push_back({toid, s->industry[toid], 1});
                }
            }
        }
    }

#ifdef DEBUG
    for (int i = 1; i <= s->count; i++)
    {
        printf("(%d,%d) ", s->cord[i].first, s->cord[i].second);
    }
    puts("");
#endif // 打印每个州的中心点

#ifdef DEBUG
    int cnt = 0;
    for (int i = 1; i <= s->count; i++)
    {
        for (edge node : s->Edge[i])
        {
            printf("%d %d %d\n ", i, node.v, node.w);
            cnt++;
        }
    }
    printf("%d %d\n", s->count, cnt);
#endif // 打印每个顶点的邻接表

#ifdef DEBUG
    for (int i = 1; i <= s->count; i++)
    {
        std::cout << s->industry[i] << " ";
    }
    puts("");
#endif // 打印每个州的发达程度

#ifdef DEBUG
    for (int i = 0; i < p->height; i++)
    {
        for (int j = 0; j < p->width; j++)
        {
            std::cout << std::setw(5) << ids[i * p->width + j] << " ";
        }
        puts("");
    }
#endif // 打印染色后的图

    delete[] ids;
}

int solve1(struct State *s)
{
    int ans = dij2(s, 1);
    return ans;
}

int solve2(struct State *s)
{
    // TODO
    if (s->pre == nullptr)
    {
        solve1(s);
    }
    int *path = new int[s->count + 1]();
    int pathid = 0;
    getpath(path, s->count, pathid, s->pre);
#ifdef DEBUG_PATH
    printf("----This is Shortest PATH----\n");
    for (int i = 0; i < pathid; i++)
    {
        printf("%d", path[i]);
        if (i != pathid - 1)
        {
            printf("->");
        }
    }
    puts("");
    printf("-----------------------------\n");
#endif
    long long ans = 0x3f3f3f3f;
    for (int i = 0; i < pathid - 1; i++)
    {
        int u = path[i];
        int v = path[i + 1];
        for (edge &e : s->Edge[u])
        {
            if (e.v == v)
            {
                e.f = 0;
            }
        }

        // int *pre = nullptr;
        long long tmp = dij2(s, 0); // ans
        // int *tmppath = new int[s->count + 1]();
        // int tmpid = 0;
        // getpath(tmppath, s->count, tmpid, pre);
#ifdef DEBUG_TMP_PATH
        printf("----This is the Second Shortest PATH----\n");
        for (int i = 0; i < tmpid; i++)
        {
            printf("%d", tmppath[i]);
            if (i != tmpid - 1)
            {
                printf("->");
            }
        }
        printf(" value:%d",tmp);
        puts("");
        printf("-----------------------------\n");
#endif
        // delete[] tmppath;
        if (ans > tmp)
        {
            if(s->mind&&tmp>s->mind)
            {
                ans = tmp;
            }
        }
        for (edge &e : s->Edge[u])
        {
            if (e.v == v)
            {
                e.f = 1;
            }
        }
    }
    delete[] path;
    return ans;
}

// long long dij3(struct State *s, int *(&pre))
// {
//     if (pre == nullptr)
//     {
//         pre = new int[s->count + 1]();
//     }
//     struct node
//     {
//         long long dis, u;

//         bool operator>(const node &a) const { return dis > a.dis; }
//     };
//     const long long inf = 0x3f3f3f3f;
//     long long *dis = new long long[s->count + 1];
//     int *vis = new int[s->count + 1]();
//     std::priority_queue<node, std::vector<node>, std::greater<node>> q;
//     for (size_t i = 0; i < s->count + 1; i++)
//     {
//         dis[i] = inf;
//     }
//     dis[1] = 0;
//     q.push({0, 1});
//     while (!q.empty())
//     {
//         int u = q.top().u;
//         q.pop();
//         if (vis[u])
//             continue;
//         vis[u] = 1;
//         for (auto ed : s->Edge[u])
//         {
//             int v = ed.v, w = ed.w;
//             if (ed.f)
//             {
//                 if (dis[v] > dis[u] + w)
//                 {
//                     dis[v] = dis[u] + w;

//                     pre[v] = u;
// #ifdef DEBUG_DIJPATH
//                     printf("%d->%d:%d\n", u, v, w);
// #endif

//                     q.push({dis[v], v});
//                 }
//             }
//         }
//     }
// #ifdef DEBUG_1
//     for (int i = 0; i < s->count; i++)
//     {
//         printf("pre[%d]:%d ", i, s->pre[i]);
//     }
// #endif
//     long long ans = dis[s->count];
//     delete[] vis;
//     delete[] dis;
//     return ans;
// }

long long dij2(struct State *s, int opt)
{
    if (s->pre == nullptr && opt)
    {
        s->pre = new int[s->count + 1]();
    }
    struct node
    {
        long long dis, u;

        bool operator>(const node &a) const { return dis > a.dis; }
    };
    const long long inf = 0x3f3f3f3f;
    long long *dis = new long long[s->count + 1];
    int *vis = new int[s->count + 1]();
    std::priority_queue<node, std::vector<node>, std::greater<node>> q;
    for (size_t i = 0; i < s->count + 1; i++)
    {
        dis[i] = inf;
    }
    dis[1] = 0;
    q.push({0, 1});
    while (!q.empty())
    {
        int u = q.top().u;
        q.pop();
        if (vis[u])
            continue;
        vis[u] = 1;
        for (auto ed : s->Edge[u])
        {
            int v = ed.v, w = ed.w;
            if (ed.f)
            {
                if (dis[v] > dis[u] + w)
                {
                    dis[v] = dis[u] + w;
                    if (opt)
                    {
                        s->pre[v] = u;
#ifdef DEBUG_DIJPATH
                        printf("%d->%d:%d\n", u, v, w);
#endif
                    }
                    q.push({dis[v], v});
                }
            }
        }
    }
#ifdef DEBUG_1
    for (int i = 0; i < s->count; i++)
    {
        printf("pre[%d]:%d ", i, s->pre[i]);
    }
#endif
    long long ans = dis[s->count];
    if(opt)
    {
        s->mind = ans;
    }
    delete[] vis;
    delete[] dis;
    return ans;
}

/// @brief 获取路径
/// @param path 路径存入path
/// @param x 从x往前推
/// @param pathid path大小存入pathid
/// @param pre 前驱结点数组
void getpath(int *path, int x, int &pathid, int *pre)
{
    if(pre==nullptr)
    {
        return;
    }
    if (x == 1)
    {
        path[pathid++] = x;
        return;
    }
    getpath(path, pre[x], pathid, pre);
    path[pathid++] = x;
}

inline bool in(int x, int y, int width, int height)
{
    return x >= 0 && x < width && y >= 0 && y < height;
}

// aid function
// 对图进行染色 用来标记id
void color(struct PNG *p, int *map, const int id, const int sx, const int sy, int &center_x, int &center_y)
{
    int dir[4][2] = {{1, 0}, {0, -1}, {-1, 0}, {0, 1}};
    int max_x = -1, max_y = -1;
    int min_x = INT_MAX, min_y = INT_MAX;
    std::queue<std::pair<int, int>> q;
    q.push({sx, sy});
    PXL *origin_pxl = get_PXL(p, sx, sy);
    map[sy * p->width + sx] = id;
    while (!q.empty())
    {
        std::pair<int, int> coordinate = q.front();
        q.pop();

        max_x = max(max_x, coordinate.first);
        max_y = max(max_y, coordinate.second);
        min_x = min(min_x, coordinate.first);
        min_y = min(min_y, coordinate.second);

        for (int i = 0; i < 4; i++)
        {
            int tx = coordinate.first + dir[i][0];
            int ty = coordinate.second + dir[i][1];
            if (in(tx, ty, p->width, p->height))
            {
                PXL *pixel = get_PXL(p, tx, ty);
                if ((*pixel == *origin_pxl) && map[ty * p->width + tx] == 0)
                {
                    map[ty * p->width + tx] = id;
                    q.push({tx, ty});
                }
            }
        }
    }
    center_x = (max_x + min_x) / 2;
    center_y = (max_y + min_y) / 2;
}

int get_id(int *ids, int x, int y, int width, int height)
{
    if (!in(x, y, width, height))
    {
        return 0;
    }
    return ids[y * width + x];
}

int cal_industry(PXL *pxl)
{
    return (255 * 255 * 3 - pxl->red * pxl->red - pxl->green * pxl->green - pxl->blue * pxl->blue);
}

int max(int a, int b)
{
    if (a > b)
    {
        return a;
    }
    return b;
}

int min(int a, int b)
{
    if (a < b)
    {
        return a;
    }
    return b;
}

2024-07-10T12:35:30.png

2024-07-10T12:35:43.png

#include <stdio.h>
#include <string.h>
#include <math.h>
#include "io.h"
#define MAXN 20

char map[MAXN][MAXN], vis[MAXN][MAXN][MAXN * MAXN][4];
int k;
int ex, ey;
int n, m;
int dir[4][2] = {{1, 0}, {0, -1}, {-1, 0}, {0, 1}};

int in(int x, int y)
{
    return (x >= 0 && x < n && y >= 0 && y < m);
}
char visnake[10];
int ans = MAXN * MAXN * 10;
int sx, sy;
// walk
int tmp = 0;
void walk(char tmpmap[MAXN][MAXN], int x, int y, int tox, int toy)
{
    if (tmpmap[tox][toy] == '0' + k)
    {
        tmp = tmpmap[tox][toy] - '0';
    }
    tmpmap[tox][toy] = tmpmap[x][y];
    if (tmpmap[x][y] == '0' + k)
    {
        tmpmap[x][y] = '.';
    }
    if ((tmp) && (tmpmap[tox][toy] == tmp - 1 + '0'))
    {
        tmpmap[x][y] = tmp + '0';
    }
    int nx = x, ny = y;
    for (int i = 0; i < 4; i++)
    {
        int tx = x + dir[i][0];
        int ty = y + dir[i][1];
        if (tmpmap[tx][ty] == tmpmap[x][y] + 1)
        {
            visnake[tmpmap[tx][ty] - '0'] = 1;
            nx = tx, ny = ty;
            break;
        }
    }
    if (nx == x && ny == y)
    {
        return;
    }
    walk(tmpmap, nx, ny, x, y);
}
// update
void update(char tmpmap[MAXN][MAXN], int sx, int sy, int tx, int ty)
{
    tmp = 0;
    memset(visnake, 0, sizeof(visnake));
    walk(tmpmap, sx, sy, tx, ty);
    //disp(tmpmap, n, m); // 每update一次就输出
}

struct node
{
    char map[MAXN][MAXN];
    int x, y;
    int step;
    int dir;
} q[MAXN * MAXN * 10];

struct node make_node(const char map[MAXN][MAXN], int x, int y, int step, int dir)
{
    struct node tmp;
    memcpy(tmp.map, map, MAXN * MAXN);
    tmp.x = x, tmp.y = y;
    tmp.step = step;
    tmp.dir = dir;
    return tmp;
}

int bfs()
{
    int l = 0, r = 0;
    for (int i = 0; i < 4; i++)
    {
        char tmp[MAXN][MAXN];
        memcpy(tmp, map, MAXN * MAXN);
        int tx = sx + dir[i][0];
        int ty = sy + dir[i][1];
        if (in(tx, ty) && !vis[tx][ty][1][i] && ((tmp[tx][ty] == '0' + k&&k!=2) || tmp[tx][ty] == '@' || tmp[tx][ty] == '.'))
        {
            vis[tx][ty][1][i] = 1;
            update(tmp, sx, sy, tx, ty);
            q[r++] = make_node(tmp, tx, ty, 1, i);
        }
    }
    while (l < r)
    {
        struct node tmp = q[l++];
        char tmpmap[MAXN][MAXN];
        memcpy(tmpmap, tmp.map, MAXN * MAXN);
        for (int i = 0; i < 4; i++)
        {
            memcpy(tmpmap, tmp.map, MAXN * MAXN);
            int tx = tmp.x + dir[i][0], ty = tmp.y + dir[i][1];
            if (in(tx, ty) && !vis[tx][ty][tmp.step + 1][i] && (((tmpmap[tx][ty] == '0' + k&&k!=2)) || (tmpmap[tx][ty] == '@') || (tmpmap[tx][ty] == '.')))
            {
                if(tmpmap[tx][ty]=='@')
                {
                    return tmp.step+1;
                }
                vis[tx][ty][tmp.step + 1][i] = 1;
                update(tmpmap, tmp.x, tmp.y, tx, ty);
                q[r++] = make_node(tmpmap, tx, ty, tmp.step + 1, i);
            }
        }
    }
    return -1;
}

void user_control()
{
    //disp(map, n, m);
    int ch;

    while (scanf("%d", &ch))
    {
        switch (ch)
        {
        case 2:
            // down
            update(map, sx, sy, sx + 1, sy);
            sx++;
            break;
        case 4:
            // left
            update(map, sx, sy, sx, sy - 1);
            sy--;
            break;
        case 8:
            // up
            update(map, sx, sy, sx - 1, sy);
            sx--;
            break;
        case 6:
            // right
            update(map, sx, sy, sx, sy + 1);
            sy++;
            break;
        default:
            break;
        }
    }
}
#define RUN
int main()
{
    // freopen("out.txt", "w+", stdout);

    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
    {
        scanf("%s", map[i]);
    }

    disp(map, n, m);
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (map[i][j] >= '1' && map[i][j] <= '9')
            {
                if (map[i][j] - '0' > k)
                {
                    k = map[i][j] - '0';
                }
            }
            if (map[i][j] == '1')
            {
                sx = i, sy = j;
            }
            if (map[i][j] == '@')
            {
                ex = i, ey = j;
            }
        }
    }
#ifdef RUN
    printf("%d", bfs());
#endif
#ifdef DEBUG
    user_control();
#endif
}

正方形 - 计蒜客

2024-07-09T12:29:56.png

题意分析:

搜索4条边,使得4条边满足相等而且等于sum/4,即数组所有元素一起能构成一个正方形

初代算法:

dfs(int id, int l1,int l2,int l3,int l4)
{
    dfs(id+1,l1+p[id],l2,l3,l4);
    dfs(id+1,l1,l2+p[id],l3,l4);
    dfs(id+1,l1,l2,l3+p[id],l4);
    dfs(id+1,l1,l2,l3,l4+p[id]);
}

对4条边进行搜索 会超时

分析:

每条边都是等效的,因此重复状态很多

不是很理解ac代码的k的作用.

#include <stdio.h>
#define MAXN 25
long long p[MAXN], n;
long long sum = 0;
long long a = 0;

int ok = 0;
long long vis[MAXN];
// x表示当前正在搜索第x(0,1,2)条边
// sum 表示当前搜索的边的长度
// k表示//避免重复状况(?)
// 即 若当前dfs的k仍从1开始,
// 遇到一个下标介于1和k之间的下标(记为j),
// 并且发现vis[j]=0,
// 那么其实已经说明j被这层dfs前面的dfs搜索过,
// 而且还不是正确答案状态,
// 可以省略这些状态,
// 所以每层k都要进行+1
void dfs(int sum,int k,int x)
{
    if(ok)
    {
        return;
    }
    if(x==3)
    {
        ok=1;
        return;
    }
    if(sum>a)
    {
        return;
    }
    if(sum==a)
    {
        dfs(0,0,x+1);
    }
    for(int i =k;i<=n;i++)
    {
        if(!vis[i])
        {
            vis[i]=1;
            dfs(sum+p[i],i+1,x);
            vis[i]=0;
        }
    }
}

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &p[i]);
        sum += p[i];
    }
    if(sum%4!=0)
    {
        printf("No");
        return 0; 
    }
    a = sum / 4;

    for(int i = 1;i<=n;i++)
    {
        if(p[i]>a)
        {
            printf("No");
            return 0;
        }
    }
    dfs(0,1,0);
    if (ok)
    {
        printf("Yes");
    }
    else
    {
        printf("No");
    }
}

引爆炸弹 - 来自计蒜客小学期

题目:

2024-07-08T04:52:04.png 2024-07-08T04:52:17.png

题意分析:

1.引爆一个炸弹,则属于该行该列的炸弹都会被引爆,结果被引爆的每个炸弹也递归引爆剩余的炸弹

2.最少需要手动引爆多少个,我们可以把会连续引爆的炸弹们归为同一个集合里面,这个集合里的任何一个炸弹被引爆,整个集合都会被引爆。因此这个集合的所有炸弹是等效的

3.那么我们只需要首先引爆某个炸弹,然后对集合进行标记引爆,搜索染色直到所有炸弹都炸了。

思路:

  1. 统计地图中的所有炸弹和炸弹坐标

  2. 先引爆某个炸弹

  3. 对这个炸弹所在的集合进行染色

  4. 深搜,即枚举剩下的炸弹,找到剩下的集合 如果找到了 则手动的步数+1

注意:

  1. 只需要进行染色即可 不需要recolor

  2. dfs起点只需要随意一个炸弹即可

  3. 地图没有存储的必要

代码:


#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAXN 666
char map[MAXN][MAXN];
int n, m;
int ans = MAXN * MAXN;
struct BOMB
{
    int x, y;
} bombs[MAXN * MAXN];
struct VISB
{
    char visb;
    int father;
} visb[MAXN * MAXN];
int bombcnt;
int exploded;
int queue[MAXN*MAXN],tot;
inline int same_block(int i, int j)
{
    if ((bombs[i].x == bombs[j].x) || (bombs[i].y == bombs[j].y))
    {
        return 1;
    }
    return 0;
}
void color_dfs(int b)
{
    queue[++tot] = b;
    while(tot!=0)
    {
        int x = queue[tot--];
        for(int i = 0;i<bombcnt;i++)
        {
            if(same_block(x,i)&&!visb[i].visb)
            {
                exploded++;
                if(exploded==bombcnt)
                {
                    return;
                }
                visb[i].visb = 1;
                visb[i].father = x;
                queue[++tot] = i;
            }
        }
    }
}
void decolor_dfs(int b)
{
    queue[++tot] = b;
    while(tot!=0)
    {
        int x = queue[tot--];
        for(int i =0;i<bombcnt;i++)
        {
            if(same_block(x,i)&&visb[i].visb&&visb[i].father==x)
            {
                visb[i].visb = 0;
                exploded--;
                visb[i].father = -1;
                queue[++tot] = i;
            }
        }
    }
}

int dfs(int b, int cnt)
{
    if (cnt > ans)
    {
        return 0;
    }
    color_dfs(b);
    if (exploded == bombcnt)
    {
        printf("%d",cnt);
        return 1;
    }
    for (int i = 0; i < bombcnt; i++)
    {
        if (!visb[i].visb)
        {
            if(dfs(i, cnt + 1))
            {
                return 1;
            }
        }
    }
    //decolor_dfs(b);
    return 0;
}
int main()
{
    freopen("E:\\Downloads\\4.txt","r",stdin);
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
    {
        scanf("%s", map[i]);
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (map[i][j] == '1')
            {
                bombs[bombcnt].x = i;
                bombs[bombcnt].y = j;
                bombcnt++;
            }
        }
    }

    dfs(rand()%bombcnt, 1);

}