HDU——1242Rescue（BFS+优先队列求点图最短路）

 

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25203    Accepted Submission(s): 8936

Problem Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

 

 

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend. 

Process to the end of the file.

 

 

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life." 

 

 

Sample Input

7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........

 

 

Sample Output

13

 

 

 

自从懂了点BFS，这些就都是水过了。

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x) memset(x,0,sizeof(x))
#define MMINF(x) memset(x,INF,sizeof(x))
typedef long long LL;
const double PI=acos(-1.0);
const int N=205;
char pos[N][N];
int vis[N][N];
int n,m;
struct info
{
	int x;
	int y;
	int time;
	bool operator<(const info &b)const
	{
		return time>b.time;
	}
};
inline info operator+(const info &a,const info &b)
{
	info c;
	c.x=a.x+b.x;
	c.y=a.y+b.y;
	return c;
}
info direct[4]={{1,0,0},{0,1,0},{-1,0,0},{0,-1,0}};
info S;
priority_queue<info>Q;
void init()
{
	MM(pos);
	MM(vis);
	while (!Q.empty())
		Q.pop();
}
bool check(const info &a)
{
	return (a.x>=0&&a.x<n&&a.y>=0&&a.y<m&&!vis[a.x][a.y]&&pos[a.x][a.y]!='#');
}
int main(void)
{
	int i,j;
	while (~scanf("%d%d",&n,&m))
	{
		init();
		for (i=0; i<n; i++)
		{
			for (j=0; j<m; j++)
			{
				cin>>pos[i][j];
				if(pos[i][j]=='a')
				{
					S.x=i;
					S.y=j;
				}
			}	
		}
		int r=-1;
		S.time=0;
		vis[S.x][S.y]=1;
		Q.push(S);
		while (!Q.empty())
		{
			info now=Q.top();
			Q.pop();
			if(pos[now.x][now.y]=='r')
			{
				r=now.time;
				break;
			}
			for (i=0; i<4; i++)
			{
				info v=now+direct[i];
				if(check(v))
				{
					v.time=now.time+(pos[v.x][v.y]=='x'?2:1);
					vis[v.x][v.y]=1;
					Q.push(v);
				}
			}
		}
		r==-1?puts("Poor ANGEL has to stay in the prison all his life."):printf("%d\n",r);
	}
	return 0;
}