hdu 4856 Tunnels(bfs+状态压缩)

题目链接：hdu 4856 Tunnels

题目大意：给定一张图，图上有M个管道，管道给定入口和出口，单向，现在有人想要体验下这M个管道，问最短需要移动的距离，起点未定。

解题思路：首先用bfs处理出两两管道之间移动的距离，然后后用状态压缩求出最短代价，dp[i][j]，i表示的已经走过的管道，j是当前所在的管道。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 20;
const int INF = 0x3f3f3f3f;
const int dir[4][2] = { {0, 1}, {1, 0}, {-1, 0}, {0, -1} };

struct point {
    int x, y;
    point (int x = 0, int y = 0) {
        this->x = x;
        this->y = y;
    }
};

struct pipe {
    point s, e;
} pi[maxn];

int N, M, d[maxn][maxn];
int dp[(1<<15)+5][maxn];
char g[maxn][maxn];

int bfs (point s, point e) {
    int vis[maxn][maxn];
    memset(vis, -1, sizeof(vis));

    queue<point> que;

    vis[s.x][s.y] = 0;
    que.push(s);

    while (!que.empty()) {
        point u = que.front();
        que.pop();

        if (u.x == e.x && u.y == e.y)
            return vis[u.x][u.y];

        for (int i = 0; i < 4; i++) {
            int x = u.x + dir[i][0];
            int y = u.y + dir[i][1];

            if (x <= 0 || x > N || y <= 0 || y > N)
                continue;

            if (vis[x][y] != -1 || g[x][y] == '#')
                continue;

            vis[x][y] = vis[u.x][u.y] + 1;
            que.push(point(x, y));
        }
    }
    return -1;
}

void init () {
    for (int i = 1; i <= N; i++)
        scanf("%s", g[i] + 1);

    memset(d, 0, sizeof(d));

    for (int i = 0; i < M; i++) {
        scanf("%d%d%d%d", &pi[i].s.x, &pi[i].s.y, &pi[i].e.x, &pi[i].e.y);

        for (int j = 0; j < i; j++) {
            d[i][j] = bfs(pi[i].e, pi[j].s);
            d[j][i] = bfs(pi[j].e, pi[i].s);
        }
    }
}

int solve () {
    memset(dp, INF, sizeof(dp));

    for (int i = 0; i < M; i++)
        dp[1<<i][i] = 0;

    for (int s = 0; s < (1<<M); s++) {

        for (int j = 0; j < M; j++) {
            if (dp[s][j] == INF)
                continue;

            for (int k = 0; k < M; k++) {
                if (s&(1<<k))
                    continue;

                if (d[j][k] == -1)
                    continue;

                dp[s|(1<<k)][k] = min(dp[s|(1<<k)][k], dp[s][j] + d[j][k]);
            }
        }
    }

    int ans = INF;
    for (int i = 0; i < M; i++)
        ans = min(ans, dp[(1<<M)-1][i]);
    return ans == INF ? -1 : ans;
}


int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        init();

        printf("%d\n", solve());;
    }
    return 0;
}