POJ-2329 Nearest number - 2（BFS）

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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");        }    }}

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