题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<queue>

using namespace std;
const int N = 1010;
int height[N][N];
bool st[N][N];
int n;

void bfs(int x, int y, bool &has_heigher, bool &has_lower)
{
    int dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, dy[8] = {1, 1, 0, -1, -1, -1, 0, 1};

    queue<pair<int, int>> q;
    q.push({x, y});
    st[x][y] = true;

    while(q.size())
    {
        auto t = q.front();
        q.pop();
        int x = t.first, y = t.second;
        for (int i = 0; i < 8; i ++)
        {
            int a = x + dx[i], b = y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= n) continue;
            if (height[a][b] != height[x][y])
            {
                if (height[a][b] > height[x][y]) has_heigher = true;
                else has_lower = true;
            }
            else if (!st[a][b])
            {
                q.push({a, b});
                st[a][b] = true;
            }
        }
    }
}

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

    int peak = 0, valley = 0;

    for (int i = 0; i < n; i ++)
        for (int j = 0; j < n; j ++)
            if (!st[i][j])
            {
                bool has_heigher = false, has_lower = false;
                bfs(i, j, has_heigher, has_lower);
                if (!has_heigher) peak ++;
                if (!has_lower) valley ++;
            }

    cout << peak << " " << valley;
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla