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

 

【题意】n*n的矩阵，对于不是0的地方仍为原值，是0的地方，找出离他最近的不是零的地方，如果有离他一样近的两个及以上地方就仍为0，就一个的话，为那一个的值。

【思路】巧用int dx[]= {1,1,-1,-1},cx[]= {-1,0,1,0};int dy[]= {1,-1,-1,1},cy[]= {0,1,0,-1};

还有判定条件if(k>n) return 0;

参考资料：http://blog.csdn.net/code_or_code/article/details/26274451

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=222;
int mp[N][N];
int n;
int dx[]= {1,1,-1,-1},cx[]= {-1,0,1,0};
int dy[]= {1,-1,-1,1},cy[]= {0,1,0,-1};
bool go(int x,int y)
{
    if(x<1||x>n||y<1||y>n) return false;
    else return true;
}
int bfs(int x,int y,int k)
{
   if(k>n) return 0;
    if(n==1||mp[x][y]) return mp[x][y];
    int cnt=0,flag=0;
    int xx,yy;
    int tmpx,tmpy;
    for(int i=0; i<4; i++)
    {
        xx=x+k*cx[i];
        yy=y+k*cy[i];
        for(int j=0; j<k; j++)
        {
            if(mp[xx][yy]&&go(xx,yy))
            {
                if(cnt==1)
                {
                    flag=1;
                    break;
                }
                cnt++;
                tmpx=xx,tmpy=yy;
            }
            xx+=dx[i];
            yy+=dy[i];

        }
        if(flag) break;
    }
    if(cnt==0) return bfs(x,y,k+1);
    else if(flag) return 0;
    else return mp[tmpx][tmpy];
}
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            scanf("%d",&mp[i][j]);
        }
    }
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            printf("%d ",bfs(i,j,1));
        }
        printf("\n");
    }
    return 0;
}