HZAU 1097 Yuchang and Zixiang ‘s maze （BFS）

1097: Yuchang and Zixiang ‘s maze

Time Limit: 2 Sec  Memory Limit: 128 MB
Submit: 863  Solved: 149
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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 45 55 74 66 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****************************************************************/