HDU-5025 2014广州网络赛 Saving Tang Monk 状压+BFS

给出一个N*N的矩阵，开启牢门需要收集齐m种钥匙，且必须收集了前i-1种钥匙才能收集第i种钥匙，最终收集齐了回到关押唐僧的房间拯救唐僧，经过一个‘S‘的房间时需要额外耗时把蛇打死，蛇最多5条，所以状压一下用优先队列BFS求最小时间即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <queue>
#include <vector>
#include <cstring>
#include <algorithm>
#define inf 1<<29
using namespace std;
const int maxn=111;
char str[maxn][maxn];
int n,m;
int ans;
int map[maxn][maxn];
int dp[maxn][maxn][11];//访问到坐标(x,y)身上有i个钥匙的步数
int dir[4][2]= {0,1,0,-1,1,0,-1,0};
struct node
{
	int x,y;
	int step;
	int snake;
	int k;
 	friend bool operator<(node a,node b)
    {
        return a.step>b.step;
    }
};
int go(int x,int y)
{
	if(x>=0&&x<n&&y>=0&&y<n&&str[x][y]!='#')
	{
		return 1;
	}
	return 0;
}
void solve(int x,int y)
{
	priority_queue<node> q;
	node front,now;
	now.x=x;
	now.y=y;
	now.step=0;
	now.snake=0;
	now.k=0;
	q.push(now);
	while(!q.empty())
	{
		front=q.top();
		q.pop();
		x=front.x;
		y=front.y;
		if(str[x][y]=='T'&&front.k==m)
		{
			ans=min(ans,front.step);
		}
		for(int i=0;i<4;i++)
		{
			int fx=x+dir[i][0];
			int fy=y+dir[i][1];
			now.x=fx;
			now.y=fy;
			now.step=front.step+1;
			now.snake=front.snake;
			now.k=front.k;
			if(go(fx,fy))
			{
				if(str[fx][fy]=='S')
				{
					int k=map[fx][fy];
					if(((1<<k)&now.snake)==0)
                    {
                        now.step++;
                        now.snake|=(1<<k);
                    }
				}
				if(str[fx][fy]-'0'==now.k+1)
				{
					now.k+=1;
				}
				if(dp[fx][fy][now.k]>now.step&&now.step<ans)
				{
					dp[fx][fy][now.k]=now.step;
					q.push(now);
				}
			}
		}
	}
}
int main()
{
	int cnt;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==0&&m==0)
		{
			break;
		}
		cnt=0;
		for(int i=0;i<n;i++)
		{
			scanf("%s",str[i]);
		}
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(str[i][j]=='S')
				{
					map[i][j]=cnt++;
				}
				for(int k=0;k<11;k++)
				{
					dp[i][j][k]=inf;
				}
			}
		}
		ans=inf;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(str[i][j]=='K')
				{
					solve(i,j);
					break;
				}
			}
		}
		if(ans==inf)
		{
			printf("impossible\n");
		}
		else
		{
			printf("%d\n",ans);
		}
	}
	return 0;
}