AcWing 173. 矩阵距离(搜索)

题目链接


题目描述

  • 请见原题目!
  • 该题目为 求某个点到离它最近的起点的距离

题目代码

/*
  求某个点到离它最近的起点的距离
  
  tt = -1 :
  其他初始为 0是因为 main里面传了一个值作为队尾
  也就是传入的 q[0]是队尾,而这没有传值
  如果 tt还是从 0开始, ++ tt就会导致 q[0] 这个位置为空,
  所以下标从 -1开始
*/
#include <cstring>
#include <iostream>
#include <algorithm>
#define x first
#define y second
using namespace std;
typedef pair<int, int> PII;
const int N = 1010, M = N * N;
int n, m;
char g[N][N];
PII q[M];
int dist[N][N];
void bfs()
{
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    memset(dist, -1, sizeof dist);
    int hh = 0, tt = -1;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            if (g[i][j] == '1')
            {
                dist[i][j] = 0;
                q[ ++ tt] = {i, j};
            }
    while (hh <= tt)
    {
        auto t = q[hh ++ ];
        for (int i = 0; i < 4; i ++ )
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 1 || a > n || b < 1 || b > m) continue;
            if (dist[a][b] != -1) continue;
            dist[a][b] = dist[t.x][t.y] + 1;
            q[ ++ tt] = {a, b};
        }
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i ++ ) scanf("%s", g[i] + 1);
    bfs();
    for (int i = 1; i <= n; i ++ )
    {
        for (int j = 1; j <= m; j ++ ) printf("%d ", dist[i][j]);
        puts("");
    }
    return 0;
}
posted @ 2022-07-12 16:50  esico  阅读(25)  评论(0)    收藏  举报