洛谷 P3957 跳房子 二分+DP检验+单调队列优化

题意：有n个格子，第i个格子距离起点xi得分si，有一个机器人只能跳d的距离，花1金币可以增加1灵活度，问要得到k分至少需要多少金币

1 ≤ n ≤ 5e5, 1 ≤ d ≤2000, 1 ≤ xi, k ≤ 1e9, |si| < 1e5

思路：花多少金币的决策有单调性，所以先二分答案，用dp检验，设dp[i]表示在第i个格子获得的最大分数

容易看出dp[i] = max(dp[k])+s[i]，k能走到i，显然不优化是n^2的复杂度

但也比较显然这就是单调队列优化的模型（P1725），于是nlogn AC

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#define INF 0x3f3f3f3f
#define LL long long
#define debug(x) cout << "[" << x << "]" << endl
using namespace std;

const int mx = 5e5+10;
int x[mx], s[mx];
LL dp[mx];
int n, k, d;

bool check(int g){
    deque<int> q;
    int p = 0, l = max(1, d-g), r = d+g;
    for (int i = 1; i <= n; i++){
        dp[i] = -INF;
        while (x[i]-x[p] >= l && p < i){
            while (!q.empty() && dp[p] >= dp[q.back()]) q.pop_back();
            q.push_back(p++);
        }
        while (!q.empty() && x[i]-x[q.front()] > r) q.pop_front();
        if (q.empty() || dp[q.front()] == -INF) continue;
        dp[i] = dp[q.front()]+s[i];
        if (dp[i] >= k) return 1;
    }
    return 0;
}

int main(){
    scanf("%d%d%d", &n, &d, &k);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &x[i], &s[i]);
    int l = 0, r = 2e9, ans = -1;
    while (l <= r){
        int mid = l+(r-l)/2;
        if (check(mid)) {
            ans = mid;
            r = mid-1;
        }
        else l = mid+1;
    }
    printf("%d
", ans);
    return 0;
}