HDU 4771 Stealing Harry Potter's Precious (生成排列＋枚举 + BFS)

#include <cstdio>
#include <cstring>
#include <queue>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
int const MAX = 100 + 5;
int map[MAX][MAX];
int n, m, k, cnt;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
char s[105];

struct Point  //存点的信息
{
    int x, y;
    int step;
    int id;
}p[105];
int ex, ey;
int dis[105][105];
bool vis[105][105];

void BFS(Point pt, Point ept)
{
    vis[pt.x][pt.y] = true;
    Point now, t;
    queue <Point> q;
    q.push(pt);
    while(!q.empty())
    {
        now = q.front();
        q.pop();
        if(now.x == ex && now.y == ey) //记录两点间的最短路
        {
            dis[pt.id][ept.id] = now.step;
            dis[ept.id][pt.id] = now.step;
            return;
        }
        for(int i = 0; i < 4; i ++)
        {
            t = now;
            t.x += dx[i];
            t.y += dy[i];
            t.step ++;
            if(t.x >= n || t.x < 0 || t.y >= m || t.y < 0 || vis[t.x][t.y] || map[t.x][t.y] == 0)
                continue;
            vis[t.x][t.y] = true;
            q.push(t);
        }
    }
}

int main()
{
    while(scanf("%d %d", &n, &m) && (n + m))
    {
        bool f = false;
        int ans = 0;
        string next;
        char tmp[10];
        cnt = 0;
        memset(map, 0, sizeof(map));
        memset(dis, -1, sizeof(dis));
        for(int i = 0; i < n; i++)
        {
            scanf("%s", s);
            for(int j = 0; j < m; j++)
            {
                if(s[j] == '.')
                    map[i][j] = 1;
                if(s[j] == '@')
                {
                    map[i][j] = 3;
                    p[cnt].x = i;
                    p[cnt].y = j;
                    p[cnt].id = cnt;
                    p[cnt++].step = 0;
                }
            }
        }
        int bx, by;
        scanf("%d", &k);
        for(int i = 0; i < k; i++)
        {
            scanf("%d %d", &bx, &by);
            if(map[bx - 1][by - 1] == 0)
                f = true;
            else
                map[bx - 1][by - 1] = 2;
        }
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                if(map[i][j] == 2)
                {
                    p[cnt].x = i;
                    p[cnt].y = j;
                    p[cnt].id = cnt;
                    p[cnt++].step = 0;
                }  //将起点设置为0点方便后面的操作
        for(int i = 0; i < cnt; i++)  //枚举任两点并搜其最短路径
        {
            for(int j = i + 1; j < cnt; j++)
            {
                ex = p[j].x;
                ey = p[j].y;
                memset(vis, false, sizeof(vis));
                BFS(p[i],p[j]);
            }
        }
        for(int i = 0; i < cnt; i++) //如果存在宝藏不可达则说明无解
        {
            for(int j = 0; j < cnt; j++)
            {
                if(i == j)
                continue;
                if(dis[i][j] == -1)
                {
                    f = true;
                    break;
                }
            }
            if(f)
                break;
        }
        if(f)
        {
            printf("-1\n");
            continue;
        }
        //计算每种情况下的总路程
        for(int i = 0; i < cnt - 1; i++)
            tmp[i] = i + 49;
        next.insert(0,tmp,0,cnt-1);
        int cur = 0;
        for(int i = 0; i < cnt - 1; i++)
        {
            ans += dis[cur][next[i] - 48];
            cur = next[i] - 48;
        }
        while(next_permutation(next.begin(), next.end())) //举出所有排列
        {
            int re = 0;
            cur = 0;
            for(int i = 0; i < cnt - 1; i++)
            {
                re += dis[cur][next[i] - 48];
                cur = next[i] - 48;
            }
            ans = min(ans, re);
        }
        printf("%d\n", ans);
    }
}