poj2329dfs

Nearest number - 2

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4262		Accepted: 1325

Description

Input is the matrix A of N by N non-negative integers. 


A distance between two elements Aij and Apq is defined as |i − p| + |j − q|.



Your program must replace each zero element in the matrix with the 
nearest non-zero one. If there are two or more nearest non-zeroes, the 
zero must be left in place.

Constraints

1 ≤ N ≤ 200, 0 ≤ Ai ≤ 1000000 

Input

Input contains the number N followed by N² integers, representing the matrix row-by-row.

Output

Output must contain N² integers, representing the modified matrix row-by-row.

Sample Input

3
0 0 0
1 0 2
0 3 0

Sample Output

1 0 2
1 0 2
0 3 0

Source

Northeastern Europe 2003, Far-Eastern Subregion

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
int n,map[210][210];
int ans[210][210],xx[4]={-1,-1,1,1},yy[4]={1,-1,-1,1};

int dfs(int x,int y)
{
    int k,a;
    for(int i=1;i<2*n;i++)//到(x,y)的距离为i 
    {
        k=0;
        if(y-i>0&&map[x][y-i]) {a=map[x][y-i];k++;}
        if(y+1<=n&&map[x][y+i]) {a=map[x][y+i];k++;}
        if(x-i>0&&map[x-i][y]) {a=map[x-i][y];k++;}
        if(x+i<=n&&map[x+i][y]) {a=map[x+i][y];k++;}
        if(k>1) return 0;
        for(int j=1;j<i;j++)//横坐标绝对值差j 
        {
            for(int p=0;p<4;p++)//左上左下右上右下 
            {
                int xa=x+xx[p]*j,ya=y+yy[p]*(i-j);
                if(xa>0&&xa<=n&&ya>0&&ya<=n&&map[xa][ya]) {a=map[xa][ya];k++;}   
                if(k>1) return 0;
            }        
        }
        if(k==1) return a;
    }
    if(k==1) return a;
    else return 0;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) scanf("%d",&map[i][j]);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(map[i][j]==0) ans[i][j]=dfs(i,j);
            else ans[i][j]=map[i][j];            
        }        
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++) printf("%d ",ans[i][j]);
        cout<<endl;
    }

    return 0;
}