HDU-4568 Hunter 状态压缩

题意：给定一个网格图，图中有一些点要求全部走到，问最少的花费是多少，从任意边界进入，任意边界出去，如果不能够全部走到，输出0。

解法：一次spfa从边界上的所有点出发，计算到K个宝藏的最短路，然后计算出任意两个宝藏之间的最短路，最后状态压缩枚举即可。

代码如下：

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

// 记得要带走全部的物品 

const int INF = 0x3f3f3f3f;
int N, M, K;
int mp[205][205];
int idis[15][15]; // 这个15*15的矩阵用来保留宝藏之间的最短路程
int odis[15];      // 从边界到K个位置的最短距离

struct Node {
    int x, y;
}p[15];

int que[1000005];
int dis[40005];
char vis[40005];
int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};

bool legal(int x, int y) {
    if (x < 0 || x >= N || y < 0 || y >= M) return false;
    return true;
}

void spfa(int sta, int num) {
    int front = 0, tail = 0;
    memset(dis, 0x3f, sizeof (dis));
    memset(vis, 0, sizeof (vis));
    que[tail++] = sta;
    dis[sta] = 0, vis[sta] = 1;
    while (front < tail) {
        int cur = que[front++], nxt;
        vis[cur] = 0;
        int x = cur / M, y = cur % M;
        int xx, yy;
        for (int k = 0; k < 4; ++k) {
            xx = x + dir[k][0], yy = y + dir[k][1];
            nxt = xx * M + yy;
            if (legal(xx, yy)) {
                if (dis[nxt] > dis[cur] + mp[xx][yy]) {
                    dis[nxt] = dis[cur] + mp[xx][yy];
                    if (!vis[nxt]) {
                        vis[nxt] = 1;
                        que[tail++] = nxt;
                    }
                }
            }
        }
    }
    for (int i = 0; i < K; ++i) {
        idis[num][i] = dis[p[i].x * M + p[i].y];
    }
}

void bspfa() {
    int front = 0, tail = 0;
    memset(dis, 0x3f, sizeof (dis));
    memset(vis, 0, sizeof (vis));
    for (int i = 0; i < N; ++i) {
        int k1 = i * M, k2 = i * M + M-1;
        que[tail++] = k1, que[tail++] = k2;
        dis[k1] = mp[i][0], dis[k2] = mp[i][M-1]; // 边界点均被初始距离为0加入进来 
        vis[k1] = vis[k2] = 1;
    }
    for (int j = 1; j < M - 1; ++j) {
        int k1 = j, k2 = (N-1) * M + j;
        que[tail++] = k1, que[tail++] = k2;
        dis[k1] = mp[0][j], dis[k2] = mp[N-1][j];
        vis[k1] = vis[k2] = 1;
    }
    while (front < tail) {
        int cur = que[front++], nxt;
        vis[cur] = 0;
        int x = cur / M, y = cur % M;
        int xx, yy;
        for (int k = 0; k < 4; ++k) {
            xx = x + dir[k][0], yy = y + dir[k][1];
            nxt = xx * M + yy;
            if (legal(xx, yy)) {
                if (dis[nxt] > dis[cur] + mp[xx][yy]) {
                    dis[nxt] = dis[cur] + mp[xx][yy];
                    if (!vis[nxt]) {
                        vis[nxt] = 1;
                        que[tail++] = nxt;
                    }
                }
            }
        }
    }
    for (int i = 0; i < K; ++i) {
        odis[i] = dis[p[i].x*M + p[i].y];
    }
}

int f[13][1<<13];
// f[i][j]表示状态为j，并且最后走的位置为i的最少开销 

int dfs(int sta, int nxt) {
    if (~f[nxt][sta] && nxt != -1) {
        return f[nxt][sta];
    }
    if (sta == 0) {
        return f[nxt][sta] = odis[nxt]; // 从nxt开始进入
    }
    int ret = INF;
    for (int i = 0; i < K; ++i) {
        if (sta&(1 << i)) {
            if (nxt != -1)
                ret = min(ret, dfs(sta^(1 << i), i) + idis[i][nxt]);
            else
                ret = min(ret, dfs(sta^(1 << i), i) + odis[i] - mp[p[i].x][p[i].y]);
                // 走i点走出去的 
        }
    }
    return f[nxt][sta] = ret;
}
 
int solve() {
    memset(f, 0x3f, sizeof (f));
    int mask = 1 << K;
    for (int i = 0; i < K; ++i) f[i][1<<i] = odis[i];
    for (int i = 2; i < mask; ++i) {
        if (!(i - (i&(-i)))) continue; // 如果只有一位为1
        for (int j = 0; j < K; ++j) {
            if (!(i&(1<<j))) continue;
            for (int k = 0; k < K; ++k) {
                f[j][i] = min(f[j][i], f[k][i^(1<<j)] + idis[k][j]);
            }
        }
    }
    int ret = INF;
    for (int i = 0; i < K; ++i) {
        ret = min(ret, f[i][mask-1] + odis[i] - mp[p[i].x][p[i].y]);
    }
    return ret;
}

int main() {
    int T;
//    freopen("1.in", "r", stdin);
    scanf("%d", &T);
    while (T--) {
        scanf("%d %d", &N, &M);
        memset(idis, 0x3f, sizeof (idis));
        memset(odis, 0x3f, sizeof (odis));
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                scanf("%d", &mp[i][j]);
                if (mp[i][j] == -1) mp[i][j] = INF;
            }
        }
        scanf("%d", &K);
        for (int i = 0; i < K; ++i) {
            scanf("%d %d", &p[i].x, &p[i].y);
        }
        for (int i = 0; i < K; ++i) {
            spfa(p[i].x * M + p[i].y, i);
        }
        bspfa();
        memset(f, 0xff, sizeof (f));
        int ret = dfs((1<<K)-1, -1);
    //    int ret = solve();   // 也可 
        if (ret == INF) puts("0");
        else printf("%d\n", ret); 
    }
    return 0;
}