（BFS）Rescue--HDOJ

Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1425 Accepted Submission(s): 558

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

Author

CHEN, Xue

Source

ZOJ Monthly, October 2003

Recommend

Eddy

总结：

需要注意的是 朋友有很多个，从每个朋友到达a的时间中取最短的一个，那么也就是说从a开始走，利用BFS遇到的第一个朋友，就是时间最短的路线

#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<math.h>
#define PI acos(-1.0)
#define eps 0.00000001

using namespace std;

typedef struct Node
{
int x,y,ti;
}Node;

char mp[205][205];
Node fri[205];
int vis[205][205];
int dirx[4] = {1,-1,0,0};
int diry[4] = {0,0,-1,1};
int tarx,tary,n,m;

int BFS(int xx,int yy)
{
queue<Node> q;
Node t;
t.x = xx,t.y = yy,t.ti=0;//这里的时间
vis[t.x][t.y] = 1;
q.push(t);
memset(vis,0,sizeof(vis));
while(!q.empty())
{
Node tmp = q.front();
q.pop();
for(int i=0; i<4; ++i)
{
int xx = tmp.x + dirx[i];
int yy = tmp.y + diry[i];
if(xx<0||xx==n||yy<0||yy==m) continue;
if(mp[xx][yy] == '#') continue;
if(vis[xx][yy]) continue;
Node nxt;
nxt.x = xx,nxt.y = yy;
if(mp[xx][yy] == '.')
nxt.ti = tmp.ti + 1;
else if(mp[xx][yy] == 'x')
nxt.ti = tmp.ti + 2;
else if(mp[xx][yy] == 'r')
{
nxt.ti = tmp.ti + 1;
return nxt.ti;
}
vis[xx][yy] = 1;
q.push(nxt);
}
}
return 999999;
}

int main(void)
{
//   freopen("in.txt","r",stdin);

while(scanf("%d %d",&n,&m) != EOF)
{
int numFri=0;
for(int i=0; i<n; ++i)
for(int j=0; j<m; ++j)
{
cin >> mp[i][j];
if(mp[i][j] == 'a')
{
tarx = i;
tary = j;
}
}
int ans = 999999;
ans = BFS(tarx,tary);

if(ans == 999999)
cout << "Poor ANGEL has to stay in the prison all his life."<<endl;
else
cout << ans <<endl;
}
return 0;
}