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Walk Out

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1977    Accepted Submission(s): 373


Problem Description
In an nm技术分享 maze, the right-bottom corner is the exit (position (n,m)技术分享 is the exit). In every position of this maze, there is either a 0技术分享 or a 1技术分享 written on it.

An explorer gets lost in this grid. His position now is (1,1)技术分享 , and he wants to go to the exit. Since to arrive at the exit is easy for him, he wants to do something more difficult. At first, he‘ll write down the number on position (1,1)技术分享 . Every time, he could make a move to one adjacent position (two positions are adjacent if and only if they share an edge). While walking, he will write down the number on the position he‘s on to the end of his number. When finished, he will get a binary number. Please determine the minimum value of this number in binary system.
 

 

Input
The first line of the input is a single integer T (T=10)技术分享 , indicating the number of testcases.

For each testcase, the first line contains two integers n技术分享 and m (1n,m1000)技术分享 . The i技术分享 -th line of the next n技术分享 lines contains one 01 string of length m技术分享 , which represents i技术分享 -th row of the maze.
 

 

Output
For each testcase, print the answer in binary system. Please eliminate all the preceding 0技术分享 unless the answer itself is 0技术分享 (in this case, print 0技术分享 instead).
 

 

Sample Input
2 2 2 11 11 3 3 001 111 101
 

 

Sample Output
111 101
 

 

Author
XJZX
 

 

Source
 

 

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题目意思:
给定一个n*m的图,图上的每个点上的值为0或1,让你找到一条路,所经过的01序列最小。
从当前点来看,如果能走0肯定要走0,如果没有0才会走1,我们可以先找到能到的最远的0,然后从最远的0处,向下和向右两个方向搜索了,
(斜行递推,从第一个行开始标记下一行要走的点,然后在下一行的时候检查那些点被标记了,说明这些点可以走,然后再标记下一行的,
直到倒数第二行,由于下一行的点要么是0,要么是1,所以可以直接输出,即使下一行有多个点可以走,但是它都是0或者1)
这样处理可以排除特殊情况
,(直接从两个方向搜,可能会漏掉往左走最终到达的这个路径上全是0的这种情况)。
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
#include <queue>
#include <queue>
#define SIZE 1005
#define maxn 2010
using namespace std;
int n,m;
char Map[SIZE][SIZE];
int  visit[SIZE][SIZE];
int  d[SIZE][SIZE]; 
int  VIS[maxn][maxn];
int tot;
int dir[4][2]={0,-1,0,1,-1,0,1,0};
struct node
{
    int x,y,cnt;
};
queue <node>  que;
void init()
{
    tot=1;
   memset(visit,0,sizeof(visit));
   memset(d,0,sizeof(d));
   memset(VIS,0,sizeof(VIS));
}
void dfs(int x,int y)
{
    if(visit[x][y]==1)
        return ;
    visit[x][y]=1;
    if(Map[x][y]==1)
        return ;
    d[x][y]=1;
    if(x+y>tot)
        tot=x+y;
    if(x>1)
        dfs(x-1,y);
    if(x<n)
        dfs(x+1,y);
    if(y>1)
        dfs(x,y-1);
    if(y<m)
        dfs(x,y+1);
}
void bfs()
{
    if(Map[1][1]==0)
    {
        node a,b,c;
        a.x=1; a.y=1; a.cnt=2;
        que.push(a);
        VIS[1][1]=1;
        d[1][1]=1;
        while(!que.empty())
        {
             b=que.front();
             que.pop();
             if(b.cnt > tot)
                tot=b.cnt;
             for(int i=0;i<4;i++)
             {
                 int next_x=b.x+dir[i][0];
                 int next_y=b.y+dir[i][1];
                 if( (1<=next_x) &&(next_x<=n) && (1<=next_y) && (next_y<=m) && VIS[next_x][next_y]==0 && Map[next_x][next_y]==0)
                 {
                     c.x=next_x; c.y=next_y;
                     c.cnt=next_x+next_y;
                     que.push(c);
                     VIS[next_x][next_y]=1;
                     d[next_x][next_y]=1;
                 }
             }

        }
    }
    else
        return ;

}
void solve()
{
    if(tot==m+n)
    {
        printf("0\n");
        return ;
    }
    if(tot==1)
    {
       tot=2; //代表起始点为(1,1)
       d[1][1]=1;
       printf("1");
    }
    for(int p=tot;p<n+m;p++)
    {
       int flag=1;
       for(int q=max(p-m,1);q<=min(p-1,n);q++)  //q代表行号
       {
           if(d[q][p-q])
           {
               int x=(Map[q][p-q+1]-0)?1:0;
               int y=(Map[q+1][p-q]-0)?1:0;
               flag=min(flag,x);
               flag=min(flag,y);
           }
       }
       for(int q=max(p-m,1);q<=min(p-1,n);q++)
       {
           if(d[q][p-q])
           {
               int x=(Map[q][p-q+1]-0)?1:0;
               int y=(Map[q+1][p-q]-0)?1:0;
               if(x==flag)
                  d[q][p-q+1]=1;
               if(y==flag)
                  d[q+1][p-q]=1;
           }
       }
       printf("%d",flag);
    }
     printf("\n");
}
int main()
{
    //freopen("test.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t --)
    {
        init();
        scanf("%d%d%*c",&n,&m);
       // printf("%d %d %d\n",n,m,tot);
        for(int i = 1 ; i <= n ; i ++)
        {
            scanf("%s",&Map[i][1]);
        }
       // getchar();
        //dfs(1,1);
        bfs();
      //  printf("tot: %d\n",tot);
        solve();
    }
    return 0;
}

 

 

hdu5335(搜索)

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原文地址:http://www.cnblogs.com/xianbin7/p/4692701.html

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