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\n", solve());

        } else

            printf("-1\n");

    }

    return 0;

}