Rescue（优先队列+bfs） Rescue

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Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31144 Accepted Submission(s): 10926

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

/**************************************************************
Problem: Rescue
Language: C/C++
method: 优先队列+bfs
date:2017 7 18
create by Songdan_Lee
****************************************************************/
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int dir[4][2]={-1,0,1,0,0,-1,0,1};
char map[205][205];
int vis[205][205];
int n,m;
typedef struct Node{
int x;
int y;
int step;
bool operator < (const Node &a)
const {
return step>a.step;//最小值优先
}
}ss;
typedef struct Node1{
int x;int y;
}position;
position start,end1;
void bfs(int x,int y,int total){
priority_queue<ss>q;
ss hd,hd1;
memset(vis,0,sizeof(vis));
vis[x][y]=1;
hd.x=x;
hd.y=y;
hd.step=total;
q.push(hd);
while(!q.empty())
{
hd1=q.top();
q.pop();
if(map[hd1.x][hd1.y]=='a')
{
printf("%dn",hd1.step+1);
return ;
}
for(int i=0;i<4;i++)
{
hd.x=hd1.x+dir[i][0];
hd.y=hd1.y+dir[i][1];
if(hd.x<0||hd.y<0||hd.x>=m||hd.y>=n||vis[hd.x][hd.y]||map[hd.x][hd.y]=='#')
continue;
if(map[hd.x][hd.y]=='.')
hd.step=hd1.step+1;
if(map[hd.x][hd.y]=='x')
hd.step+hd.step+2;
vis[hd.x][hd.y]=1;
q.push(hd);
}
}
printf("Poor ANGEL has to stay in the prison all his life.n");
}
int main(){
while(scanf("%d%d",&m,&n)!=EOF){
if(m==0&&n==0)
break;
for(int i=0;i<m;i++)
scanf("%s",map[i]);
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
if(map[i][j]=='a')
{
end1.x=i;
end1.y=j;
}
if(map[i][j]=='r'){
start.x=i;
start.y=j;
}
}
bfs(start.x,start.y,0);
}
}