HDU1242——Rescue【BFS】

最新推荐文章于 2019-02-13 22:25:45 发布

原创最新推荐文章于 2019-02-13 22:25:45 发布 · 170 阅读

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本文深入探讨了广度优先搜索(BFS)算法的基本原理及其在解决复杂路径寻找问题中的应用。通过一个具体的囚犯救援问题，展示了如何利用BFS算法高效地找到从起点到终点的最短路径。文章不仅提供了详细的算法步骤，还附带了完整的代码实现，帮助读者理解并掌握BFS算法的实际操作。

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Rescue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36992 Accepted Submission(s): 12845

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

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BFS裸题，直接上模版套就可以了。

#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define INF 1000000
using namespace std;
const int MAXN=200+10;
int sx,sy,ex,ey;//起始点，终点坐标
int n,m;
char mp[MAXN][MAXN];
int d[MAXN][MAXN];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
typedef pair<int,int> P;
void BFS(){
    queue<P> q;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++) d[i][j]=INF;
    }
    q.push(P(sx,sy));
    d[sx][sy]=0;
    while(!q.empty()){
        P p=q.front();q.pop();
        if(p.first==ex&&p.second==ey) break;
        for(int i=0;i<4;i++){
            int nx=p.first+dx[i],ny=p.second+dy[i];
            if(nx>=0&&nx<n&&ny>=0&&ny<m&&mp[nx][ny]!='#'&&d[nx][ny]==INF){
                if(mp[nx][ny]=='.'||mp[nx][ny]=='r') d[nx][ny]=d[p.first][p.second]+1;
                else d[nx][ny]=d[p.first][p.second]+2;
                q.push(P(nx,ny));
            }
        }
    }
    if(d[ex][ey]==INF) printf("Poor ANGEL has to stay in the prison all his life.\n");
    else printf("%d\n",d[ex][ey]);
}
int main(){
    while(~scanf("%d%d",&n,&m)){
        for(int i=0;i<n;i++) scanf("%s",mp[i]);
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mp[i][j]=='a'){
                    sx=i;
                    sy=j;
                }
                if(mp[i][j]=='r'){
                    ex=i;
                    ey=j;
                }
            }
        }
        BFS();
    }
    return 0;
}