落谷1607

#include<bits/stdc++.h>
using namespace std;#define ls u<<1
#define rs u<<1|1const int N = 1e5;int n, k, c, ans;
struct line
{int l, r, m; // m为候车的牛的数量bool operator<(line b) {return r < b.r;}
}s[N];struct tree
{int l, r, mx, lazy;
}t[N*4];void pushup(int u) {t[u].mx = max(t[ls].mx, t[rs].mx);
}void pushdown(int u) {if (t[u].lazy) {t[ls].mx += t[u].lazy;t[rs].mx += t[u].lazy;t[ls].lazy += t[u].lazy;t[rs].lazy += t[u].lazy;t[u].lazy = 0;}
}void build(int u, int l, int r) {// 建树t[u] = { l,r };if (l == r) return;int mid = l + r >> 1;build(ls, l, mid);build(rs, mid + 1, r);pushup(u);
}void change(int u, int l, int r, int v) {if (l > t[u].r || r < t[u].l) return;if (l <= t[u].l && r >= t[u].r) {// 覆盖的情况t[u].mx += v;t[u].lazy += v;return;}pushdown(u);change(ls, l, r, v);change(rs, l, r, v);pushup(u);
}int query(int u, int l, int r) {if (l > t[u].r || r < t[u].l) return 0;if (l <= t[u].l && r >= t[u].r) {return t[u].mx;}pushdown(u);return max(query(ls, l, r), query(rs, l, r));
}int main() {cin >> k >> n >> c;for (int i = 1; i <= k; i++) {cin >> s[i].l >> s[i].r >> s[i].m;}sort(s + 1, s + k + 1);build(1, 1, n);for (int i = 1; i <= k; i++) {int l = s[i].l, r = s[i].r;int mx = query(1, l, r - 1);int x = min(c - mx, s[i].m);change(1, l, r - 1, x);ans += x;}cout << ans;return 0;
}