hdu 4771 Stealing Harry Potter's Precious(bfs)

题目链接：hdu 4771 Stealing Harry Potter's Precious

题目大意：在一个N*M的银行里，贼的位置在’@‘，如今给出n个宝物的位置。如今贼要将全部的宝物拿到手。问最短的路径，不须要考虑离开。

解题思路：由于宝物最多才4个，加上贼的位置，枚举两两位置，用bfs求两点距离，假设存在两点间不能到达，那么肯定是不能取全然部的宝物。

然后枚举取宝物的顺序。维护ans最小。

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

using namespace std;
const int maxn = 100;
const int maxv = 10;
const int INF = 0x3f3f3f3f;
const int dir[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };

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

char g[maxn+5][maxn+5];
int n, N, M, v[maxn+5][maxn+5], d[maxv][maxv];

void init () {

    memset(g, 0, sizeof(g));

    for (int i = 1; i <= N; i++) {
        scanf("%s", g[i] + 1);
        for (int j = 1; j <= M; j++)
            if (g[i][j] == '@') {
                p[0].x = i;
                p[0].y = j;
            }
    }

    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &p[i].x, &p[i].y);
}

int bfs (point s, point e) {

    queue<point> que;
    memset(v, -1, sizeof(v));

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

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

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

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

            if (xi > N || xi <= 0)
                continue;

            if (yi > M || yi <= 0)
                continue;

            if (v[xi][yi] != -1 || g[xi][yi] == '#')
                continue;

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

bool judge () {
    memset(d, 0, sizeof(d));

    for (int i = 0; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            int dis = bfs(p[i], p[j]);

            if (dis == -1)
                return false;
            d[i][j] = d[j][i] = dis;
        }
    }
    return true;
}

int solve () {
    int pos[maxv];
    for (int i = 0; i <= n; i++)
        pos[i] = i;

    int ans = INF;
    do {
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += d[pos[i]][pos[i+1]];
        ans = min (ans, sum);
    } while(next_permutation(pos + 1, pos + n + 1));

    return ans;
}

int main () {
    while (scanf("%d%d", &N, &M) == 2 && N + M) {
        init();
        if (judge()) {
            printf("%d
", solve());
        } else
            printf("-1
");
    }
    return 0;
}