hdu 5335 Walk Out

Walk Out

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 200    Accepted Submission(s): 32

Problem Description

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

An explorer gets lost in this grid. His position now is  (1,1) , 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) . 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) , indicating the number of testcases. 

For each testcase, the first line contains two integers  n  and  m (1≤n,m≤1000) . The  i -th line of the next  n  lines contains one 01 string of length  m , which represents  i -th row of the maze.

 

Output

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

 

Sample Input

  

   2
2 2
11
11
3 3
001
111
101
  

 

Sample Output

  

   111
101
  

 

Source

2015 Multi-University Training Contest 4

先dfs出最远的0点 再从最远的对角线开始扫，一直到最后一个点

#include <bits/stdc++.h>
using namespace std;
char g[1005][1005];
int vis[1005][1005];
char ans[5005];
int top=0;
int dirx[4]= {0,0,-1,1};
int diry[4]= {1,-1,0,0};
int n,m;
bool dp[1005][1005];
int legal(int x,int y)
{
    if(x<1||x>n||y<1||y>m)
        return 0;
    return 1;
}
int mn;
queue < pair<int,int> > q;
void bfs(int x,int y)
{
    while(!q.empty())
        q.pop();
    q.push(make_pair(x,y));
    while(!q.empty())
    {
        int x=q.front().first;
        int y=q.front().second;
        q.pop();
        if(vis[x][y]<1000009)
            continue;
        if(g[x][y]=='1')
        {
            vis[x][y]=n-x+m-y;
            mn=min(mn,n-x+m-y);
            continue;
        }
        vis[x][y]=n-x+m-y;
        mn=min(mn,n-x+m-y);
        for(int k=0; k<4; k++)
            if(legal(x+dirx[k],y+diry[k]))
                q.push(make_pair(x+dirx[k],y+diry[k]));
    }
}
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        memset(vis,0x3f,sizeof(vis));
        mn=1000000009;
        n=m=1000;
        scanf("%d %d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            scanf(" %s",g[i]+1);
        }
        bfs(1,1);
        if(mn==0&&g[n][m]=='0')
        {
            puts("0");
            continue;
        }

        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
                if(vis[i][j]==mn)
                    dp[i][j]=1;
        mn--;
        top=0;
        ans[top++]='1';
        while(mn>=0)
        {
            int flag=1;
            for(int j=m; j>=1; j--)
            {
                int i=n+m-j-mn;
                if(i<1||i>n)
                    continue;
                if(i==1)
                {
                    if(dp[i][j-1]==0)
                        continue;
                }
                else if(j==1)
                {
                    if(dp[i-1][j]==0)
                        continue;
                }
                else if(dp[i-1][j]==0&&dp[i][j-1]==0)
                    continue;
                if(g[i][j]=='0')
                    flag=0;
            }
            if(flag)
            {
                for(int j=m; j>=1; j--)
                {
                    int i=n+m-j-mn;
                    if(i<1||i>n)
                        continue;
                    if(i==1)
                    {
                        if(dp[i][j-1]==0)
                            continue;
                    }
                    else if(j==1)
                    {
                        if(dp[i-1][j]==0)
                            continue;
                    }
                    else if(dp[i-1][j]==0&&dp[i][j-1]==0)
                        continue;
                    dp[i][j]=1;
                }
            }
            else
            {
                for(int j=m; j>=1; j--)
                {
                    int i=n+m-j-mn;
                    if(i<1||i>n)
                        continue;
                    if(i==1)
                    {
                        if(dp[i][j-1]==0)
                            continue;
                    }
                    else if(j==1)
                    {
                        if(dp[i-1][j]==0)
                            continue;
                    }
                    else if(dp[i-1][j]==0&&dp[i][j-1]==0)
                        continue;
                    if(g[i][j]=='0')
                        dp[i][j]=1;
                }
            }
            mn--;
            ans[top++]='0'+flag;
        }
        for(int i=0; i<top; i++)
            printf("%c",ans[i]);
        puts("");
    }
    return 0;
}