HDU 5335 Walk Out

Walk Out

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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
  

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 4

/*************************************************************/

题意：一个由字符'0'、'1'组成的n*m的迷宫，起点在左上角(1,1)，终点在右下角(n,m)，每次探险者可以往上下左右四个方向移动，并且记下当前位置的字符。到达终点后，他会得到一个二进制数(即探险者记录下来的字符串)，要求探险者如何走能得到最小的二进制数，输出该二进制数(不含前导0，若字符串每个字符都为'0'，则输出'0')。

解题思路：

      广搜(BFS)+贪心，要使得到的二进制数最小，而且不包括前导0，那么我们很容易就可以想到遇到1之前都走0，走到距离终点最近的位置(即与终点曼哈顿距离最小的点，亦或者说x+y(x,y为当前位置的坐标)尽可能大的点)使得后面的路径尽可能短。

①先是bfs找到离目标最近的距离；

②后用贪心的思想让靠前的字符尽可能为0。

策略是每往前走一步，判断是否 可以是0，有0就走0，没0才走1，直到终点。

给几组测试数据

100
5 5
01101
00011
01111
11111
11110

100
1 1
1

100

1 1

0


100
5 5
10011
11010
00111
11111
00010


100
5 5
00011
11011
10011
11011
00001

闲话不多说，上AC代码，用时140Ms

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 1005;
const int inf = 1000000000;
char s[N][N];
bool v[N][N];
int g[4][2]={{0,1},{1,0},{0,-1},{-1,0}},n,m,a,b;
struct node
{
    int x,y;
    node(){}
    node(int x1,int y1):x(x1),y(y1){}
};
void BFS(int x,int y)
{
    memset(v,false,sizeof(v));
    v[x][y]=true;
    if(s[x][y]=='1')
    {
        a=b=1;
        return ;
    }
    int i,Max=0;
    queue<node> q;
    q.push(node(x,y));
    while(!q.empty())
    {
        for(i=0;i<4;i++)
        {
            x=q.front().x+g[i][0];
            y=q.front().y+g[i][1];
            if(!v[x][y]&&x>0&&y>0&&x<=n&&y<=m&&s[x][y]=='0')
            {
                v[x][y]=true;
                q.push(node(x,y));
            }
        }
        if(q.front().x+q.front().y>Max)
            a=q.front().x,b=q.front().y,Max=q.front().x+q.front().y;
        q.pop();
    }
}
int main()
{
	int t,i,j,k,l,x,y;
	bool flag;
	scanf("%d",&t);
	while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            scanf("%s",s[i]+1);
        BFS(1,1);
        if(n+m==2)
        {
            printf("%c\n",s[1][1]);
            continue;
        }
        else if(a+b==n+m)
        {
            puts("0");
            continue;
        }
       // printf("%d %d\n",a,b);
        if(a+b==2)
            printf("1");
        for(i=a+b;i<n+m;i++)
        {
            for(j=1;j<=n;j++)
            {
                k=i-j;
                flag=false;
                if(k>0&&v[j][k])
                {
                    for(l=0;l<2;l++)
                    {
                        x=j+g[l][0];
                        y=k+g[l][1];
                        if(x>0&&y>0&&x<=n&&y<=m&&s[x][y]=='0')
                            flag=true;
                    }
                }
                if(flag)
                    break;
            }
            if(flag)
                printf("0");
            else
                printf("1");
            for(j=1;j<=n;j++)
            {
                k=i-j;
                if(k>0&&v[j][k])
                {
                    for(l=0;l<2;l++)
                    {
                        x=j+g[l][0];
                        y=k+g[l][1];
                        if(x>0&&y>0&&x<=n&&y<=m&&(flag&&s[x][y]=='0'||!flag))
                            v[x][y]=true;
                    }
                }
            }
        }
        puts("");
    }
	return 0;
}

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