[BZOJ4636]蒟蒻的数列

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
	if(Head == Tail) {
		int l = fread(buffer, 1, BufferSize, stdin);
		Tail = (Head = buffer) + l;
	}
	return *Head++;
}
int read() {
	int x = 0, f = 1; char c = Getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = Getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = Getchar(); }
	return x * f;
}

#define maxn 40010
#define oo 2147483647
#define LL long long
int n, lc[maxn], rc[maxn], root, Cur;
struct Node {
	int x[2], mx[2], mn[2], val, maxv;
	bool operator < (const Node& t) const { return x[Cur] != t.x[Cur] ? x[Cur] < t.x[Cur] : x[Cur^1] < t.x[Cur^1]; }
} ns[maxn], x;

void maintain(int o) {
	int l = lc[o], r = rc[o];
	for(int i = 0; i < 2; i++) {
		ns[o].mx[i] = max(max(ns[l].mx[i], ns[r].mx[i]), ns[o].x[i]);
		ns[o].mn[i] = min(min(ns[l].mn[i], ns[r].mn[i]), ns[o].x[i]);
	}
	ns[o].maxv = max(max(ns[l].maxv, ns[r].maxv), ns[o].val);
	return ;
}
void build(int& o, int L, int R, int cur) {
	if(L > R){ o = 0; return ; }
	int M = L + R >> 1; o = M;
	Cur = cur; nth_element(ns + L, ns + M, ns + R + 1);
	build(lc[o], L, M - 1, cur ^ 1); build(rc[o], M + 1, R, cur ^ 1);
	maintain(o);
	return ;
}
bool all(int o) { return ns[o].mx[0] <= x.x[0] && ns[o].mn[1] >= x.x[1]; }
bool has(int o) { return ns[o].mn[0] <= x.x[0] && ns[o].mx[1] >= x.x[1]; }
int query(int o) {
	if(!o) return 0;
	int l = lc[o], r = rc[o], ans = 0;
	if(ns[o].x[0] <= x.x[0] && ns[o].x[1] >= x.x[1]) ans = ns[o].val;
	if(ns[l].maxv > ns[r].maxv) {
		if(all(l)) ans = max(ans, ns[l].maxv);
		else if(has(l) && ns[l].maxv > ans) ans = max(ans, query(l));
		if(all(r)) ans = max(ans, ns[r].maxv);
		else if(has(r) && ns[r].maxv > ans) ans = max(ans, query(r));
	}
	else {
		if(all(r)) ans = max(ans, ns[r].maxv);
		else if(has(r) && ns[r].maxv > ans) ans = max(ans, query(r));
		if(all(l)) ans = max(ans, ns[l].maxv);
		else if(has(l) && ns[l].maxv > ans) ans = max(ans, query(l));
	}
	return ans;
}

struct Cmd { int a, b, k; } cs[maxn];
int cnt, num[maxn<<2], tp[maxn<<2];
LL ans;
int main() {
	ns[0].mx[0] = ns[0].mx[1] = -oo;
	ns[0].mn[0] = ns[0].mn[1] = oo;
	ns[0].maxv = -oo;
	n = read();
	for(int i = 1; i <= n; i++) {
		num[++cnt] = cs[i].a = read();
		num[++cnt] = cs[i].a + 1;
		num[++cnt] = cs[i].b = read() - 1;
		num[++cnt] = cs[i].b + 1;
		cs[i].k = read();
	}
	sort(num + 1, num + cnt + 1);
	cnt = unique(num + 1, num + cnt + 1) - num;
//	for(int i = 1; i <= cnt; i++) printf("%d ", num[i]); putchar(‘\n‘);
	for(int i = 1; i <= n; i++) {
		int x0 = lower_bound(num + 1, num + cnt + 1, cs[i].a) - num,
			x1 = lower_bound(num + 1, num + cnt + 1, cs[i].b) - num;
		ns[i].x[0] = x0; tp[x0] = 1;
		ns[i].x[1] = x1; tp[x1] = 1;
		ns[i].maxv = ns[i].val = cs[i].k;
	}
	for(int i = 1; i <= cnt; i++) if(!tp[i]) tp[i] = num[i+1] - num[i-1] - 1;
	
//	for(int i = 1; i <= n; i++) printf("%d: %d %d %d\n", i, ns[i].x[0], ns[i].x[1], ns[i].val);
	build(root, 1, n, 0);
//	for(int i = 1; i <= n; i++) printf("%d: %d %d %d\n", i, ns[i].x[0], ns[i].x[1], ns[i].val);
	for(int i = 1; i <= cnt; i++) {
		x.x[0] = x.x[1] = i;
		int val = query(root);
//		printf("%d: %d %d\n", i, val, tp[i]);
		ans += (LL)tp[i] * (LL)val;
	}
	
	printf("%lld\n", ans);
	
	return 0;
}