【BZOJ4636】蒟蒻的數列【掃描線】【set】

【題目鏈接】

從Claris那裏學來的掃描線做法。

/* Think Thank Thunk */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>

using namespace std;

typedef long long LL;

const int maxn = 80005;

int n, m;
LL ans;

struct data {
	int p, k;

	bool operator < (const data &x) const {
		return p != x.p ? p < x.p : k < x.k;
	}
} line[maxn];

multiset<int> S;

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

int main() {
	n = iread();
	for(int i = 1; i <= n; i++) {
		int l = iread(), r = iread(), k = iread();
		if(l >= r) continue;
		line[++m] = (data){l, k};
		line[++m] = (data){r, -k};
	}
	sort(line + 1, line + 1 + m);
	S.insert(0);
	for(int i = 1; i < m; i++) {
		if(line[i].k > 0) S.insert(line[i].k);
		else S.erase(S.find(-line[i].k));
		ans += (LL)(line[i + 1].p - line[i].p) * *S.rbegin();
	}
	printf("%lld\n", ans);
	return 0;
}