HDOJ - 1242 Rescue

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我是靠谱客的博主纯情毛巾，最近开发中收集的这篇文章主要介绍HDOJ - 1242 Rescue，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Rescue

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

一脸懵逼。。。就这么看着“学姐”的代码敲完了。。代码还是很好懂的，主要就是细节

！

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m,ans;
int a,b,c,d;
char map[205][205];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
struct node{
int x,y,time;
};
bool operator<(const node &x,const node &y){
return x.time>y.time;
}
int bfs(){
int i;
node now,next;
now.x=a;now.y=b;now.time=0;
priority_queue<node> q;
q.push(now);
while(!q.empty()){
now=q.top();
q.pop();
if(now.x==c&&now.y==d)
return now.time;
for(i=0;i<4;i++){
next.x=now.x+dx[i];
next.y=now.y+dy[i];
if(next.x>=0&&next.y>=0&&next.x<n&&next.y<m&&map[next.x][next.y]!='#'){
if(map[next.x][next.y]=='.'||map[next.x][next.y]=='a')
next.time=now.time+1;
else
next.time=now.time+2;
map[next.x][next.y]='#';
q.push(next);
}
}
}
return -1;
}
int main(){
while(~scanf("%d%d",&n,&m)){
for(int i=0;i<n;i++){
getchar();
for(int j=0;j<m;j++){
scanf("%c",&map[i][j]);
if(map[i][j]=='r')
a=i,b=j;
if(map[i][j]=='a')
c=i,d=j;
}
}
ans=bfs();
if(ans==-1)
printf("Poor ANGEL has to stay in the prison all his life.n");
else
printf("%dn",ans);
}
return 0;
}