HDU3681 Prison Break【状压】

题意：给一张图，机器人从F出发，关闭所有Y，问它的最小电池储量，图中G可以把电池充满

思路：把图中的F和G、Y拿出来用作状压DP的点，点之间最短距离跑下bfs。二分答案，用状压DP判断下这个储量能不能完成任务，若F点不能到达所有Y就输出-1

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;const int maxn = 355;int dis[maxn][maxn],dir[4][2]={0,1,0,-1,1,0,-1,0};int num[maxn][maxn],x[maxn],y[maxn],cnt,goal;int vis[maxn][maxn],n,m;int dp[1<<15][16];char s[maxn][maxn];void init(){memset(dis,0x3f,sizeof dis);memset(num,-1,sizeof num);goal = 0;}void bfs(){queue<int> q;for(int i = 0; i < cnt; i++)  //st point{memset(vis,0,sizeof vis);while(!q.empty())q.pop();q.push(x[i]*m+y[i]);vis[ x[i] ][ y[i] ] = 1;int time = 1;while(!q.empty()){int size = q.size();while(size--){int z = q.front();q.pop();int zx = z / m;int zy = z % m;for(int j = 0; j < 4; j++){int nx = zx + dir[j][0];int ny = zy + dir[j][1];if(nx >= 0 && nx < n && ny >= 0 && ny < m){if(s[nx][ny] != 'D' && !vis[nx][ny]){vis[nx][ny] = 1;q.push(nx*m+ny);if(num[nx][ny] != -1)dis[i][ num[nx][ny] ] = time;}}}}time++;}}}int sol(int sum){memset(dp,-1,sizeof dp);dp[1][0] = sum;int en = (1<<cnt)-1;for(int sh = 1; sh < en; sh++){for(int j = 0; j < cnt; j++){if(dp[sh][j] <= 0) continue;for(int k = 0; k < cnt; k++){if(sh & (1<<k)) continue;if(dis[j][k] > dp[sh][j]) continue;if( s[x[k]][y[k]] == 'G')dp[sh | (1<<k)][k] = sum;elsedp[sh | (1<<k)][k] = max( dp[sh][j] - dis[j][k] , dp[sh | (1<<k)][k] );if(((sh|(1<<k)) & goal) == goal)return 1;}}}return 0;}int main(void){int i,j,k;while(scanf("%d%d",&n,&m)!=EOF && n+m){init();for(i = 0; i < n; i++)scanf("%s",s[i]);cnt = 1;for(i = 0; i < n; i++){for(j = 0; j < m; j++){if(s[i][j] == 'F'){x[0] = i;y[0] = j;num[i][j] = 0;}else if(s[i][j] == 'G'){x[cnt] = i;y[cnt] = j;num[i][j] = cnt++;}else if(s[i][j] == 'Y'){x[cnt] = i;y[cnt] = j;goal |= 1<<cnt;num[i][j] = cnt++;}}}bfs();int flag = 1;for(i = 1; i < cnt; i++){if(s[ x[i] ][ y[i] ] == 'Y' && dis[0][i] == 0x3f3f3f3f){flag = 0;break;}}if(flag){int l = 0,r = n*m,mid;while(l < r){mid = (l+r) / 2;if(sol(mid))r = mid;elsel = mid+1;}printf("%d\n",l);}elseprintf("-1\n");}return 0;}