POJ-2329 Nearest number - 2（BFS）

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 <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <queue>

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<n;i++)
            for(int j=0;j<n;j++)
            {scanf("%d",&a[i][j]);b[i][j]=a[i][j];}
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                if(a[i][j]==0)
                {
                    init();
                    bfs(i,j);
                }

            }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(j!=n-1)
                    printf("%d ",b[i][j]);
                else
                    printf("%d",b[i][j]);
            }
            printf("\n");
        }
    }
}