一个很显然的观察：选择的三角形两两重叠面积为 \(0\)，否则合并更优。
考虑 dp，设 \(f_i\) 为删完 \(x_j \ge i\) 的所有点的最小花费。转移就枚举选择的三角形直角边长 \(l\)，那么 \(f_i = \min(f_{i + 1} + \sum\limits_{x_p = i} c_p, \min\limits_l f_{i + l} + \sum\limits_{i \le x_p < i + l \land y_p < k - i - l} c_p)\)，就是把三角形下方的那坨矩形的点的 \(c_i\) 算进去。
考虑直接线段树维护后面那坨式子的最小值，设 \(g_j = \sum\limits_{i \le x_p < j \land y_p < k - j} c_p\)，\(i + 1 \to i\) 时，对于每个 \(x_p = i\) 的点，让 \(g_{i + 1 \sim k - y_p - 1}\) 加上 \(c_p\) 即可。
时间复杂度 \(O((n + k) \log k)\)。
code
// Problem: E. Tenzing and Triangle
// Contest: Codeforces - CodeTON Round 5 (Div. 1 + Div. 2, Rated, Prizes!)
// URL: https://codeforces.com/contest/1842/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 200100;
const ll inf = 0x3f3f3f3f3f3f3f3fLL;
ll n, m, K, f[maxn], g[maxn];
vector<pii> vc[maxn];
struct node {
ll x, y, z;
} a[maxn];
namespace SGT {
ll tree[maxn << 2], tag[maxn << 2];
inline void pushup(int x) {
tree[x] = min(tree[x << 1], tree[x << 1 | 1]);
}
inline void pushdown(int x) {
if (!tag[x]) {
return;
}
tree[x << 1] += tag[x];
tree[x << 1 | 1] += tag[x];
tag[x << 1] += tag[x];
tag[x << 1 | 1] += tag[x];
tag[x] = 0;
}
void build(int rt, int l, int r) {
tree[rt] = inf;
if (l == r) {
return;
}
int mid = (l + r) >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
}
void update(int rt, int l, int r, int ql, int qr, ll x) {
if (ql > qr) {
return;
}
if (ql <= l && r <= qr) {
tree[rt] += x;
tag[rt] += x;
return;
}
pushdown(rt);
int mid = (l + r) >> 1;
if (ql <= mid) {
update(rt << 1, l, mid, ql, qr, x);
}
if (qr > mid) {
update(rt << 1 | 1, mid + 1, r, ql, qr, x);
}
pushup(rt);
}
void modify(int rt, int l, int r, int x, ll y) {
if (l == r) {
tree[rt] = y;
return;
}
pushdown(rt);
int mid = (l + r) >> 1;
(x <= mid) ? modify(rt << 1, l, mid, x, y) : modify(rt << 1 | 1, mid + 1, r, x, y);
pushup(rt);
}
}
void solve() {
scanf("%lld%lld%lld", &n, &K, &m);
for (int i = 1; i <= n; ++i) {
scanf("%lld%lld%lld", &a[i].x, &a[i].y, &a[i].z);
vc[a[i].x].pb(a[i].y, a[i].z);
g[a[i].x] += a[i].z;
}
SGT::build(1, 0, K);
mems(f, 0x3f);
f[K] = 0;
SGT::modify(1, 0, K, K, K * m);
for (int i = K - 1; ~i; --i) {
f[i] = f[i + 1] + g[i];
for (pii p : vc[i]) {
SGT::update(1, 0, K, i + 1, K - p.fst - 1, p.scd);
}
f[i] = min(f[i], SGT::tree[1] - m * i);
SGT::modify(1, 0, K, i, f[i] + m * i);
}
printf("%lld\n", f[0]);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}