BZOJ4636: 蒟蒻的数列(动态开节点线段树)
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2023-04-05 17:47:40
题意 "题目链接" Sol 直接上动态开节点线段树 因为只有一次询问,所以中途不需要下传标记 cpp include define LL long long using namespace std; const int MAXN = 8e6 + 10, INF = 1e9 + 10; templat ......
题意
sol
直接上动态开节点线段树
因为只有一次询问,所以中途不需要下传标记
#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\n", query(rt, 1, inf, 0)); return 0; }
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