hdu 5335 Walk Out (2015 Multi-University Training Contest 4)

Walk Out

                                                                        Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                               Total Submission(s): 194    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

 

题目大意：

       从(1,1)到(n,m)，路径形成的二进制数最大。

解题思路：

      BFS+贪心，先是bfs找到离目标近期的距离。后用贪心让最前面的尽可能为0。策略就是每往前走一步，推断是否

能够是0，方法就是找它前面离它近期的能够取0的那一位，推断能否够从那个位置走到当前的位置。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=1000+100;
char s[maxn][maxn];
bool vis[maxn][maxn];
int dir[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
struct node
{
    int x;
    int y;
};
int ans;
int n,m;
queue<node> q;
vector<node> son[maxn*2];
void BFS1()
{
    memset(vis,false,sizeof(vis));
    node p;
    p.x=1;
    p.y=1;
    ans=0;
    vis[1][1]=true;
    while(!q.empty())
        q.pop();
    if(s[1][1]=='0')
    {
        ans=2;
        q.push(p);
    }
    while(!q.empty())
    {
        node p2;
        p=q.front();
        q.pop();
        if(p.x==n&&p.y==m)
        {
            if(s[n][m]=='0')
            {
                ans=n+m;
                vis[n][m]=true;
            }
            break;
        }
        for(int i=0; i<4; i++)
        {
            p2.x=p.x+dir[i][0];
            p2.y=p.y+dir[i][1];
            if(p2.x>0&&p2.x<=n&&p2.y>0&&p2.y<=m&&!vis[p2.x][p2.y]&&s[p2.x][p2.y]=='0')
            {
                vis[p2.x][p2.y]=true;
                if(p2.x+p2.y>ans)
                {
                    ans=p2.x+p2.y;
                }
                q.push(p2);
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            scanf("%s",s[i]+1);
        }
        BFS1();
        if(ans==n+m)
            printf("0\n");
        else
        {
            for(int i=0; i<=n+m; i++)
                son[i].clear();
            int cur=0;//记录前一位0出现的位置
            if(ans==0)//起点为1
                ans=1;
            else
            {
                for(int i=1; i<=n; i++)//找到全部的离(n,m)近期的点。
                {
                    int j=ans-i;
                    if(j>=1&&j<=m&&vis[i][j]&&s[i][j]=='0')
                    {
                        node v;
                        v.x=i;
                        v.y=j;
                        son[ans].push_back(v);
                    }
                }
                cur=ans;
            }
            for(int i=ans+1; i<=n+m; i++)//枚举每一步
            {
                if(cur==0)//前面不存在0
                {
                    for(int j=1; j<=n; j++)
                    {
                        int k=i-j;
                        node v;
                        if(k>=1&&k<=m&&s[j][k]=='0')
                        {
                            v.x=j;
                            v.y=k;
                            son[i].push_back(v);
                            cur=i;
                        }
                    }
                }
                else
                {
                    for(int j=1; j<=n; j++)
                    {
                        int k=i-j;
                        node v;
                        if(k>=1&&k<=m&&s[j][k]=='0')
                        {
                            for(int l=0; l<son[cur].size(); l++)
                            {
                                v=son[cur][l];
                                if(v.x<=j&&v.y<=k&&i-cur>=j+k-v.x-v.y)//推断前面的0是否可达
                                {
                                    node w;
                                    w.x=j;
                                    w.y=k;
                                    son[i].push_back(w);
                                    break;
                                }
                            }
                        }
                    }
                    if(son[i].size()>0)
                    cur=i;
                }

            }
            for(int i=ans+1; i<=n+m; i++)
            {
                if(son[i].size()>0)
                    printf("0");
                else
                    printf("1");
            }
            printf("\n");
        }
    }
    return 0;
}