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.<br><br>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.<br><br>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.)<br>
Input
First line contains two integers stand for N and M.<br><br>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. <br><br>Process to the end of the file.<br>
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." <br>
Sample Input
7 8<br>#.#####.<br>#.a#..r.<br>#..#x...<br>..#..#.#<br>#...##..<br>.#......<br>........<br>
Sample Output
13<br>
Author
CHEN, Xue
Source
ZOJ Monthly, October 2003
題意：最少時間拯救天使，有路 牆 守衛，上下左右走，一步一分鐘，殺個守衛一分鐘
思路：一開始很容易想BFS，後來看了別人的還要用優先隊列
感想：對於優先隊列總是不能完全理解運用
AC代碼：
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
struct node
{
int x,y,step;
friend bool operator<(node n1,node n2)
{
return n2.step<n1.step;
}
};
int n,m,vis[205][205];
char mapp[205][205];
int x1,x2,y1,y2;
int to[4][2] = {1,0,-1,0,0,1,0,-1};
int check(int x,int y)
{
if(x<0 || y<0 || x>=n || y>=m || !vis[x][y] || mapp[x][y] == '#')
return 1;
return 0;
}
int bfs()
{
int i;
priority_queue<node> Q;
node a,next;
a.x = x1;
a.y = y1;
a.step = 0;
Q.push(a);
vis[x1][y1] = 0;
while(!Q.empty())
{
a = Q.top();
Q.pop();
if(a.x == x2 && a.y == y2)
return a.step;
for(i = 0; i<4; i++)
{
next = a;
next.x+=to[i][0];
next.y+=to[i][1];
if(check(next.x,next.y))
continue;
next.step++;
if(mapp[next.x][next.y] == 'x')
next.step++;
if(vis[next.x][next.y]>=next.step)
{
vis[next.x][next.y] = next.step;
Q.push(next);
}
}
}
return 0;
}
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
for(i = 0; i<n; i++)
{
scanf("%s",mapp[i]);
for(j = 0; mapp[i][j]; j++)
{
if(mapp[i][j] == 'r')
{
x1 = i;
y1 = j;
}
else if(mapp[i][j] == 'a')
{
x2 = i;
y2 = j;
}
}
}
memset(vis,1,sizeof(vis));
int ans = 0;
ans = bfs();
if(ans)
printf("%d\n",ans);
else
printf("Poor ANGEL has to stay in the prison all his life.\n");
}
return 0;
}