51nod 1281 二分+DP

题意：n个数字，只有既大于左边又大于右边的数字是山顶，要在山顶上插k面旗子，相邻旗子的距离>=k，最大化k

数据范围： n <= 50000

思路：最大化最小值，显然是二分，重点在如何写check函数

首先肯定是把所有山顶标记起来，设dp[i]为前i个点最多插多少旗子

然后：1.对于一个山顶 dp[i] = max(dp[i], dp[i-k]+1)  插旗子

2. 对于一个不是山顶 dp[i] = max(dp[i], dp[i-1]) 继承前面的值

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

const int mx = 50010;
int dp[mx], n, a[mx];
bool vis[mx];

bool check(int k){
    memset(dp, 0, sizeof dp);
    if (k == 1){
        for (int i = 1; i <= n; i++)
            if (vis[i]) return 0;
        return 1;
    }
    for (int i = 2; i <= n-1; i++){
        dp[i] = vis[i];
        if (!vis[i]) dp[i] = max(dp[i-1], dp[i]);
        else if (i >= k) dp[i] = max(dp[i], dp[i-k]+1);
    }
    return dp[n-1] < k;
}

int main(){
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 2; i <= n-1; i++){
        if (a[i] > a[i-1] && a[i] > a[i+1])
            vis[i] = 1;
    }
    int l = 1, r = n;
    while (l <= r){
        int mid = (l+r)>>1;
        if (check(mid)) r = mid-1;
        else l = mid+1;
    }
    printf("%d\n", r);
    return 0;
}