Nearest number - 2

Time Limit: 5000MS Memory Limit: 65536K

Total Submissions: 4100 Accepted: 1275

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

有DP的方法，效率是O(n^2)，但是我想不出，也没看到有博客是用DP，所以就暴力了。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;int a[205][205];int b[205][205];int vis[205][205];int dir[4][2]={ { 1,0},{ 0,1},{-1,0},{ 0,-1}};int n;int c[405];int d[405];struct Node{ int x,y;};void bfs(int x,int y){ Node Q[40005]; int rear=0,front=0; int flag=0,level=9999999; //Queue.push(Node(x,y)); Node term; term.x=x; term.y=y; Q[rear++]=term; vis[x][y]=1; while(rear!=front&&flag!=-1) { // Node temp=Queue.front(); //Queue.pop(); Node temp=Q[front++]; if(level<=vis[temp.x][temp.y]) break; for(int i=0;i<4;i++) { int xx=temp.x+dir[i][0]; int yy=temp.y+dir[i][1]; if(xx<0||xx>n-1||yy<0||yy>n-1||vis[xx][yy]) continue; vis[xx][yy]=vis[temp.x][temp.y]+1; if(a[xx][yy]!=0) { if(!flag) { flag=a[xx][yy]; level=vis[xx][yy]; } else { flag=-1; break; } } //Queue.push(Node(xx,yy)); else { Node item; item.x=xx; item.y=yy; Q[rear++]=item; } } } if(flag>0) b[x][y]=flag;}void init(){ memset(vis,0,sizeof(vis)); memset(c,0,sizeof(c));}int main(){ while(scanf("%d",&n)!=EOF) { for(int i=0;i