POJ 2329 (暴力+搜索bfs）

Nearest number - 2

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3943		Accepted: 1210

Description

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

A distance between two elements A_ij and A_pq 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

Sample Output

1 0 2
1 0 2
0 3 0

#include<iostream>
#include<cstdio>

using namespace std;

int n;
int matri[210][210];
int dx[]={1,1,-1,-1},cx[]={-1,0,1,0};
int dy[]={-1,1,1,-1},cy[]={0,-1,0,1};

bool in_matrix(int x,int y)
{
    if(x<0||x>=n) return false;
    if(y<0||y>=n) return false;
    return true;
}

int bfs(int x,int y,int k)
{
    if(k>n) return 0;                           //n*n matrix搜索K次，自己能够特值来理解
    if(matri[x][y]||n==1) return matri[x][y];   //数不为0，或仅仅有一个数（即 1*1 矩阵），就输出
    int xx,yy,X,Y;
    int i,j;
    int cnt=0,die=0;
    for(i=0;i<4;i++)           //对于菱形4条边的搜索，这里是以每边K个数字来写。
    {
        xx=x+k*cx[i];
        yy=y+k*cy[i];
        for(j=k;j--;)              //相当于for(j=0;j<k;j++),一边k个数，所以搜索k次
        {
            if(in_matrix(xx,yy)&&matri[xx][yy])
            {
                if(cnt==1)
                {
                    die=1;
                    break;
                }

                X=xx;
                Y=yy;
                cnt++;

            }
            xx+=dx[i];
            yy+=dy[i];
        }
        if(die)
            break;
    }
    if(cnt==0)
        return bfs(x,y,k+1);
    else if(die)
        return 0;
    else
        return matri[X][Y];

}



int main()
{
    scanf("%d",&n);
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j)
            scanf("%d",&matri[i][j]);

    for(int i = 0; i < n; ++i,printf("
"))
        for(int j = 0; j < n; ++j)
            printf("%d ",bfs(i,j,1));

    return 0;
}

（借鉴大大的思路）

值得学习的是，对于矩阵的逆时针菱形搜索。思考了非常长时间都没有想清楚。

自己能够试一下顺时针，一样的道理哦。

int dx[]={1,1,-1,-1},cx[]={-1,0,1,0};
int dy[]={-1,1,1,-1},cy[]={0,-1,0,1};

主要是这两对数组。用的非常是巧妙！