矩阵距离

题目描述

原文理解

曼哈顿距离

                                                               投影到坐标轴长度的和

、

                                                          红色的线为曼哈顿距离

题解

 

#include <cstring>
#include <iostream>

#define x first
#define y second

using namespace std;

typedef pair<int, int> pii;

const int N = 1010;

int n, m;
int dist[N][N];
pii q[N * N];
char g[N][N];

void bfs()
{
    int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};

    memset(dist, -1, sizeof dist);
    int tt = 0, hh = 0;
    for( int i = 0; i < n; i ++ )
        for( int j = 0; j < m; j ++ )
            if( g[i][j] == '1' )
            {
                dist[i][j] = 0;
                q[tt ++] = {i, j};
            }

    while( hh < tt )
    {
        pii cur = q[hh ++];
        int x = cur.x, y = cur.y;
        for( int i = 0; i < 4; i ++ )
        {
            int nx = x + dx[i], ny = y + dy[i];
            if( nx < 0 || nx >= n || ny < 0 || ny >= m )    continue;
            if( dist[nx][ny] != -1 )    continue;

            dist[nx][ny] = dist[x][y] + 1;
            q[tt ++] = {nx, ny};
        }
    }
}

int main()
{
    cin >> n >> m;
    for( int i = 0; i < n; i ++ )   cin >> g[i];

    bfs();

    for( int i = 0; i < n; i ++ )
    {
        for( int j = 0; j < m; j ++ )
            cout << dist[i][j] << ' ';
        cout << endl;
    }

    return 0;
}