zoj 1649 Rescue

救援行动：BFS算法实战

最新推荐文章于 2017-10-12 12:19:52 发布

原创最新推荐文章于 2017-10-12 12:19:52 发布 · 837 阅读

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#zoj bfs

本文介绍了一个经典的BFS（宽度优先搜索）算法应用场景——救援任务。在该任务中，需要利用BFS算法找到从指定起点到达被囚禁角色位置的最短路径，同时考虑了地图中的障碍物及敌人等复杂因素。

Rescue

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

//裸的bfs
#include <stdio.h>
#include <queue>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#define maxn 210
#define MM 10000000
using namespace std;
int n,m;
int ax,ay;
int xx[]={1,-1,0,0};
int yy[]={0,0,1,-1};
int mintime[maxn][maxn];
struct point
{
    int x;  //方格的位置
    int y;
    int step; //走的步数
    int time; //花的时间
}; queue<point> Q;
point start,end;
char ww[maxn][maxn];  //走到每个位置所花的最少时间。
void bfs(point p) // 从开始位置开始bfs
{
    Q.push(p);
    point t;
    while(!Q.empty())
    {
        t=Q.front();
        Q.pop();
        for(int i=0;i<4;i++)
        {     
             point tt;  //向第i个方向走一步后的位置
            int x=t.x+xx[i],y=t.y+yy[i];
            if(x<n && x>=0 && m>y && y>=0 && ww[x][y]!='#')
            {

                tt.x=x;
                tt.y=y;
                 tt.step=t.step+1;
                  tt.time=t.time+1;
                if(ww[x][y]=='x' || ww[x][y]=='X')  //杀死禁卫多花一个时间
                    tt.time++;

            }
            if(tt.time<mintime[x][y])
            {
                mintime[x][y]=tt.time;
                Q.push(tt);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
       for(int i=0;i<n;i++)
       for(int j=0;j<m;j++)
       mintime[i][j]=MM;
        for(int i=0;i<n;i++)
            scanf("%s",ww[i]);
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
              if(ww[i][j]=='r')
            {
            start.x=i;
            start.y=j;
            start.step=0;
            start.time=0;
            mintime[i][j]=0;
            }
            else if(ww[i][j]=='a')
                ax=i,ay=j;
            bfs(start);
            if(mintime[ax][ay]<MM)
            printf("%d\n",mintime[ax][ay]);
            else
                printf("Poor ANGEL has to stay in the prison all his life.\n");
    }
    return 0;
}
/*
7 8
#.#####.
#.a#.xr.
#.x#xx..
.x#.xx##
#...##..
.#......
........
*/