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
广度优先搜索找最短时间。
看代码凝视吧
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
struct node
{
int x,y,time;//x,y方格的位置。time当前全部的时间
friend bool operator<(node a,node b)//优先队列依照时间大小排序
{
return a.time>b.time;
}
};
priority_queue<node>s;
char map[205][205];//地图
int m,n,vis[205][205],escape,dir[4][2]={0,1,0,-1,1,0,-1,0};//vis标记，dir表示四个方向
bool judge(int x,int y)//推断当前位置时候能够走
{
if(x>=0&&y>=0&&x<n&&y<m&&!vis[x][y]&&map[x][y]!='#')
return true;
else
return false;
}
int tonum(int x,int y)//把对应的道路，警卫换算成时间
{
if(map[x][y]=='x')
return 2;
else
return 1;
}
void bfs(int x,int y)//广度优先搜索
{
node temp,temp1;
temp.time=0,temp.x=x,temp.y=y;
vis[x][y]=1;
s.push(temp);//把temp入队列
while(!s.empty())
{
temp=s.top(),s.pop();
temp1=temp;
if(map[temp.x][temp.y]=='a')//提前结束循环，由于用的优先队列。所以当前找到的肯定是最小的
{
escape=temp.time;
break;
}
for(int i=0;i<4;i++)
{
int xx=temp.x+dir[i][0];
int yy=temp.y+dir[i][1];
if(judge(xx,yy))
{
vis[xx][yy]=1;//当前位置已浏览 标记为1
temp.x=xx,temp.y=yy,temp.time=temp.time+tonum(xx,yy);
s.push(temp);//更新队列
}
temp=temp1;//由于才推断了一个方向。所以还须要temp1保持上次的位置
}
}
}
int main()
{
while(scanf("%d %d",&n,&m)!=EOF)
{
escape=0;
memset(vis,0,sizeof(vis));
memset(map,0,sizeof(map));
for(int i=0;i<n;i++)
scanf("%s",map[i]);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
if(map[i][j]=='r')
{
bfs(i,j);
break;
}
if(escape)
printf("%d
",escape);
else
printf("Poor ANGEL has to stay in the prison all his life.
");
while(!s.empty())//一定要记得清队列。。刚刚wa了一次
s.pop();
}
}