题目链接：点击打开链接

题意：

给定n*m的矩阵：

F：起点（有且仅有一个）

D:坏点（不能走到这个点）

G:能量池（走到这个点可以选择使用这个点的能量池，把电池充满，也可以暂时不用，只能使用一次）

Y:目标点

问：

遍历所有Y点需要最小的电池容量是多少。

开始电池满电，每走一步消耗一格电。

Y+G的个数<=15. n,m<=15

思路：状压YG，前面几位表示Y，后面几位表示G。

先跑个floyd，然后二分电池容量，状压dp判可行

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
#include <cmath>
using namespace std;
const int inf = 5500;
const int N = 15;
struct node{
    int x, y;
    node(int a=0, int b=0):x(a),y(b){}
}Y[N], G[N], P[N+1], F;
int ynum, gnum, dp[1<<N][15];
int dis[N][N][N][N], step[4][2] = {0,1, 0,-1, 1,0, -1,0};
int n, m;
char mp[N+10][N+10];
void bfs(int x, int y){
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            dis[x][y][i][j] = inf;
    dis[x][y][x][y] = 0;
    queue<int> qx, qy; qx.push(x); qy.push(y);
    while(!qx.empty()){
        int ux = qx.front(); qx.pop();
        int uy = qy.front(); qy.pop();
        for(int i = 0; i < 4; i++)
        {
            int vx = step[i][0] + ux, vy = step[i][1] + uy;
            if(false == (0<=vx && vx <n && 0<=vy && vy<m) || mp[vx][vy] == 'D')
                continue;
            if(dis[x][y][vx][vy] == inf)
            {
                dis[x][y][vx][vy] = dis[x][y][ux][uy]+1;
                qx.push(vx); qy.push(vy);
            }
        }
    }
}
void floyd(){
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            if(mp[i][j]!='D')
                bfs(i,j);
}

bool ok(int power){
    memset(dp, -1, sizeof dp);
    queue<int> State, End;
    for(int i = 0; i < ynum; i++)
        if(power - dis[F.x][F.y][Y[i].x][Y[i].y] >= 0)
        {
            dp[1<<i][i] = power - dis[F.x][F.y][Y[i].x][Y[i].y];
            State.push(1<<i); End.push(i);
        }
    for(int i = 0; i < gnum; i++)
        if(power - dis[F.x][F.y][G[i].x][G[i].y] >= 0)
        {
            dp[1<<(i+ynum)][i+ynum] = power;
            State.push(1<<(i+ynum)); End.push(i+ynum);
        }
    while(!State.empty())
    {
        int s = State.front(); State.pop();
        int e = End.front(); End.pop();
        if((s & ((1<<ynum)-1)) >= ((1<<ynum)-1))return true;
        for(int i = 0; i < ynum +gnum; i++)
        {
            if((s&(1<<i)) || dp[s][e] - dis[P[e].x][P[e].y][P[i].x][P[i].y] < 0)continue;
            if(dp[s|(1<<i)][i] == -1)
            {
                State.push(s|(1<<i));
                End.push(i);
            }
            if(i<ynum)
                dp[s|(1<<i)][i] = max(dp[s|(1<<i)][i], dp[s][e] - dis[P[e].x][P[e].y][P[i].x][P[i].y]);
            else
                dp[s|(1<<i)][i] = power;
        }
    }
    return false;
}
void input(){
    ynum = gnum = 0;
    for(int i = 0; i < n; i++)
    {
        scanf("%s", mp[i]);
        for(int j = 0; j < m; j++)
        {
            if(mp[i][j] == 'F')
                F = node(i,j);
            else if(mp[i][j] == 'Y')
                Y[ynum++] = node(i,j);
            else if(mp[i][j] == 'G')
                G[gnum++] = node(i,j);
        }
    }
    for(int i = 0; i < ynum; i++)
        P[i] = Y[i];
    for(int i = ynum; i < ynum+gnum; i++)
        P[i] = G[i-ynum];
}
int main(){
    while(cin>>n>>m, n+m){
        input();
        if(ynum == 0){puts("0");continue;}
        floyd();
        int ans = inf, l = 0, r = inf;
        while(l <= r){
            int mid = (l+r)>>1;
            if(ok(mid))
                r = mid-1, ans = min(ans, mid);
            else
                l = mid+1;
        }
        if(ans == inf) ans = -1;
        cout<<ans<<endl;
    }
    return 0;
}
/*
1 2
FY
1 4
FYGY


*/

HDU 3681 Prison Break floyd+状压+二分