HZAU 1097 Yuchang and Zixiang ‘s maze （BFS）

最新推荐文章于 2017-09-01 15:04:42 发布

原创最新推荐文章于 2017-09-01 15:04:42 发布 · 562 阅读

0 ·

CC 4.0 BY-SA版权

------题解------ 专栏收录该内容

179 篇文章

订阅专栏

本文介绍了一种使用广度优先搜索(BFS)解决大型迷宫寻路问题的方法，并提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1097: Yuchang and Zixiang ‘s maze

Time Limit: 2 Sec Memory Limit: 128 MB
Submit: 863 Solved: 149
[ Submit][ Status][ Web Board]

Description

One day , Yuchang and Zixiang go out of school to find some thing interesting . But both of them is LuChi , so the miss the way easily .

“Where am I ? who am I ?” Yuchang says . “Who attack you? You must want to say .” Zixiang adds . “Don’t say that , how we get out there ? I want my mom 555555……” Yuchang started crying . Zixiang become very panic . He doesn’t know how to get out there.

Now , they find they are in a N*M maze , they are at the (a,b) point , they know they want to go to the point (c,d) , they want to finish as soon as possible, so , could you help them ?

Same as other maze , there are some point has boom , means they can’t get the point . Give you N,M and Num (the number of points that have booms . ) , then , Num lines contains pairs (x,y) means point (x,y) have booms . then , one line contains a , b , c , d ,the begin and end point .

They can mov forward , back , left and right . And every move cost 1 second . Calculate how many seconds they need to get to the finish point .

Input

The first line contains tow numbers N,M (0 < x,y < 1000)means the size of the maze.

The second line contains a number Num (0 < N < X*Y), means the number of points which have booms .

Then next N lines each contain two numbers , xi,yi , means (xi,yi) has a boom .

Output

One line , contains one number , the time they cost .

If they can’t get to the finish point , output -1 .

Sample Input

1000 1000 4
5 5
5 7
4 6
6 6 
1 1 5 6

Sample Output

-1

题意：

典型的走迷宫问题。

小地图（200*200一下）BFS、DFS都可以。大地图的话，因为信息量很大所以用不了DFS会妥妥地T掉，只能用BFS。

代码:

#include <cstdlib>  
#include <iostream>  
#include <cstdio>  
#include <algorithm>  
#include <cmath>  
#include <vector>
#include <cstring>  
#include <queue>  
using namespace std;  
const int maxn=1001;  
const int INF=8000000;  
int save[maxn][maxn];  
int d[maxn][maxn];  
int ex,ey;  
int sx,sy;  
int move_x[4]={0,0,1,-1};  
int move_y[4]={1,-1,0,0};  
int n,m;  
int bfs(){  
    queue<pair<int, int> > que;  
    int i,j;  
    for(i=1;i<=n;i++){  
        for(j=1;j<=m;j++){  
            d[i][j]=INF;  
        }  
    }  
    que.push(make_pair(sx, sy));  
    d[sx][sy]=0;  
    while(que.size()){  
        pair<int, int> q=que.front();  
        que.pop();  
        if(q.first==ex&&q.second==ey){  
            return d[q.first][q.second];  
        }  
        for(i=0;i<4;i++){  
            int dx=q.first+move_x[i];  
            int dy=q.second+move_y[i];  
            if(dx>=1&&dx<=n&&dy>=1&&dy<=m&&save[dx][dy]!=1&&d[dx][dy]==INF){  
                que.push(make_pair(dx, dy));  
                 d[dx][dy]=d[q.first][q.second]+1;  
            }  
        }  
    }  
    return INF;  
}  
int main(){  
    while(~scanf("%d%d",&n,&m)){
        int num;
        memset(save,0,sizeof(save));
        scanf("%d",&num);
        while(num--)
        {
            int x1,y1;
            scanf("%d%d",&x1,&y1);
            save[x1][y1]=1;
        }  
        scanf("%d%d%d%d",&sx,&sy,&ex,&ey);
        int result=bfs();  
        if(result==INF){  
            printf("-1\n");  
        }else{  
            printf("%d\n",result);  
        }  
    }  
       
    return 0;  
}  
/**************************************************************
    Problem: 1097
    User: 2016_hhu09
    Language: C++
    Result: Accepted
    Time:1323 ms
    Memory:9336 kb
****************************************************************/