Hdoj 1242 Rescue

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概述

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

Author
CHEN, Xue

Source
ZOJ Monthly, October 2003
题目分析
r可能会有很多个但是a一定只有一个，从a点开始bfs就可以了，注意干掉一个guard是+2，要用优先队列
Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cctype> 
#include<cstring>
#include<cstdlib>
#include<queue> 
using namespace std;
const int dx[4]={0,-1,0,1};
const int dy[4]={-1,0,1,0};
struct node
{
int x,y,step;
friend bool operator < (const node &a,const node &b)
{
return a.step>b.step;
}
};
node st,ed,now,next;
int n,m,ans;
char map[201][201];
bool vis[201][201];
bool judge(int x,int y)
{
if(x>0 && x<=n && y>0 && y<=m && map[x][y]!='#')
return true;
return false;
}
int bfs()
{
node now,next;
priority_queue<node>q;
memset(vis,false,sizeof(vis));
if(st.x==ed.x && st.y==ed.y) return 0;
q.push(st);
vis[st.x][st.y]=true;
while(!q.empty())
{
now=q.top();
q.pop();
if(map[now.x][now.y]=='r') return now.step;
for(int i=0;i<4;i++)
{
next.x=now.x+dx[i];
next.y=now.y+dy[i];
if(judge(next.x,next.y))
{
if(!vis[next.x][next.y])
{
if(map[next.x][next.y]=='.' || map[next.x][next.y]=='r') next.step=now.step+1;
else next.step = now.step+2;
vis[next.x][next.y]=true;
q.push(next);
}
}
}
}
return -1;
}
int main()
{
while(cin>>n>>m)
{
memset(vis,false,sizeof(vis));
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
cin>>map[i][j];
if(map[i][j]=='a')
{
st.x = i; st.y = j; st.step = 0;
}
}
ans = bfs();
if(ans==-1)
cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
else cout<<ans<<endl;
}
return 0;
}