Codeforces 739C Alyona and towers 线段树

Alyona and towers

这个题写起来真的要人命。。。

我们发现一个区间被加上一个d的时候， 内部的结构是不变的， 改变的只是左端点右端点的值， 这样就能区间合并了。

如果用差分的话会简单一些， 就变成了求前一段是负数，后一段是正数的最长段多长。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 3e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, m;

LL lazy[N << 2];
struct Node {
    LL lv, rv;
    int mx, rup, rdn, lup, ldn;
    bool up, dn, mon;
    void print() {
        puts("");
        printf("lv: %lld  rv: %lld
", lv, rv);
        printf("mx: %d
", mx);
        printf("rup: %d  rdn: %d  lup: %d  ldn: %d
", rup, rdn, lup, ldn);
        printf("up: %d  dn: %d  mon: %d
", up, dn, mon);
        puts("");
    }
} a[N << 2];

Node operator + (const Node& a, const Node& b) {
    Node c;
    c.lv = a.lv; c.rv = b.rv;
    c.mx = max(a.mx, b.mx);
    c.rup = b.rup;
    c.rdn = b.rdn;
    c.lup = a.lup;
    c.ldn = a.ldn;
    
    if(a.rv < b.lv) c.mx = max(c.mx, a.rup + b.lup);
    if(a.rv > b.lv) c.mx = max(c.mx, a.rdn + b.ldn);
    if(a.rv != b.lv) c.mx = max(c.mx, a.rup + b.ldn);

    if(b.up && a.rv < b.lv) c.rup = max(c.rup, a.rup + b.rup);

    if(b.mon && a.rv < b.lv) c.rdn = max(c.rdn, a.rup + b.rdn);
    if(b.dn && a.rv > b.lv) c.rdn = max(c.rdn, a.rdn + b.rdn);

    if(a.mon && a.rv > b.lv) c.lup = max(c.lup, a.lup + b.ldn);
    if(a.up && a.rv < b.lv) c.lup = max(c.lup, a.lup + b.lup);

    if(a.dn && a.rv > b.lv) c.ldn = max(c.ldn, a.ldn + b.ldn);

    c.up = a.up && b.up && a.rv < b.lv;
    c.dn = a.dn && b.dn && a.rv > b.lv;
    c.mon = false;
    if(c.up || c.dn) c.mon = true;
    else {
        if(a.up && b.dn && a.rv != b.lv) c.mon = true;
        if(a.mon && b.dn && a.rv > b.lv) c.mon = true;
        if(b.mon && a.up && a.rv < b.lv) c.mon = true;
    }
    return c;
}

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
void push(int rt) {
    if(lazy[rt]) {
        a[rt << 1].lv += lazy[rt]; a[rt << 1].rv += lazy[rt];
        a[rt << 1 | 1].lv += lazy[rt]; a[rt << 1 | 1].rv += lazy[rt];
        lazy[rt << 1] += lazy[rt];
        lazy[rt << 1 | 1] += lazy[rt];
        lazy[rt] = 0;
    }
}

void build(int l, int r, int rt) {
    if(l == r) {
        int x; scanf("%d", &x);
        a[rt] = Node{x, x, 1, 1, 1, 1, 1, 1, 1, 1};
        return;
    }
    int mid = l + r >> 1;
    build(lson); build(rson);
    a[rt] = a[rt << 1] + a[rt << 1 | 1];
}

void update(int L, int R, int val, int l, int r, int rt) {
    if(l >= L && r <= R) {
        a[rt].lv += val; a[rt].rv += val;
        lazy[rt] += val;
        return;
    }
    int mid = l + r >> 1;
    push(rt);
    if(L <= mid) update(L, R, val, lson);
    if(R > mid)  update(L, R, val, rson);
    a[rt] = a[rt << 1] + a[rt << 1 | 1];
}

int main() {
    scanf("%d", &n);
    build(1, n, 1);
    scanf("%d", &m);
    while(m--) {
        int L, R, d;
        scanf("%d%d%d", &L, &R, &d);
        update(L, R, d, 1, n, 1);
        printf("%d
", a[1].mx);
    }
    return 0;
}

/*
*/