AcWing - 329 - 围栏障碍训练场 = 线段树 + dp

最新推荐文章于 2025-06-27 14:28:10 发布

转载最新推荐文章于 2025-06-27 14:28:10 发布 · 207 阅读

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原文链接：http://www.cnblogs.com/Inko/p/11600135.html

本文介绍了一种使用区间树和动态规划解决复杂平台跳跃问题的方法。通过从下至上反推，找到每个平台可能达到的最低能量消耗，并利用区间树进行高效更新和查询。适用于ACM竞赛及算法挑战。

https://www.acwing.com/problem/content/description/331/

貌似只能够从下往上反推，从上往下不知道走哪个方向好。每次找出这个平台A下落到哪个平台B，再从B的左右端点向上转移。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 200001;

short st[MAXN * 4 + 5], lazy[MAXN * 4 + 5];

inline void PushDown(int o) {
    if(lazy[o]) {
        lazy[o << 1] = lazy[o];
        st[o << 1] = lazy[o];
        lazy[o << 1 | 1] = lazy[o];
        st[o << 1 | 1] = lazy[o];
        lazy[o] = 0;
    }
}

void Update(int o, int l, int r, int ql, int qr, int v) {
    if(ql <= l && r <= qr) {
        lazy[o] = v;
        st[o] = v;
        return;
    }
    PushDown(o);
    int mid = l + r >> 1;
    if(ql <= mid)
        Update(o << 1, l, mid, ql, qr, v);
    if(qr >= mid + 1)
        Update(o << 1 | 1, mid + 1, r, ql, qr, v);
}

int Query(int o, int l, int r, int x) {
    if(l == r)
        return st[o];
    PushDown(o);
    int mid = l + r >> 1;
    if(x <= mid)
        return Query(o << 1, l, mid, x);
    return Query(o << 1 | 1, mid + 1, r, x);
}

const int P = 100001;

int dpL[30005], dpR[30005];
int L[30005], R[30005];

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif
    int n, s;
    scanf("%d%d", &n, &s);
    s += P;
    L[0] = P, R[0] = P;
    for(int i = 1; i <= n; ++i) {
        scanf("%d%d", &L[i], &R[i]);
        L[i] += P, R[i] += P;
        int ql = Query(1, 1, MAXN, L[i]);
        dpL[i] = min(dpL[ql] + abs(L[i] - L[ql]), dpR[ql] + abs(L[i] - R[ql]));
        int qr = Query(1, 1, MAXN, R[i]);
        dpR[i] = min(dpL[qr] + abs(R[i] - L[qr]), dpR[qr] + abs(R[i] - R[qr]));
        Update(1, 1, MAXN, L[i], R[i], i);
    }
    printf("%d\n", min(dpL[n] + abs(s - L[n]), dpR[n] + abs(s - R[n])));
}

转载于:https://www.cnblogs.com/Inko/p/11600135.html