BZOJ4636: 蒟蒻的數列(動態開節點線段樹)

自為風月馬前卒發表於2019-01-03

題意

題目連結

Sol

直接上動態開節點線段樹

因為只有一次詢問，所以中途不需要下傳標記

#include<bits/stdc++.h> 
#define LL long long 
using namespace std;
const int MAXN = 8e6 + 10, INF = 1e9 + 10;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < `0` || c > `9`) {if(c == `-`) f = -1; c = getchar();}
    while(c >= `0` && c <= `9`) x = x * 10 + c - `0`, c = getchar();
    return x * f;
}
int N, rt, ls[MAXN], rs[MAXN], mx[MAXN], tot;
void IntMax(int &k, int l, int r, int ll, int rr, int val) {
    if(!k) k = ++tot;
    if(ll <= l && r <= rr) {chmax(mx[k], val); return ;}
    int mid = l + r >> 1;
    if(ll <= mid) IntMax(ls[k], l, mid, ll, rr, val);
    if(rr  > mid) IntMax(rs[k], mid + 1, r, ll, rr, val);
}
LL Query(int k, int l, int r, int val) {
    chmax(mx[k], val); chmax(val, mx[k]);
    if(l == r) return mx[k];
    int mid = l + r >> 1;LL ans = 0;
    if(ls[k]) ans += Query(ls[k], l, mid, val);
    else ans += (mid - l + 1) * mx[k];
    if(rs[k]) ans += Query(rs[k], mid + 1, r, val);
    else ans += (r - mid) * mx[k];
    return ans;
}
signed main() {
    N = read();
    for(int i = 1; i <= N; i++) {
        int l = read(), r = read() - 1, k = read();
        IntMax(rt, 1, INF, l, r, k);
    }
    printf("%lld
", Query(rt, 1, INF, 0));
    return 0;
}

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