poj 1201

树状数组或者线段树这些数据结构只是工具，题目的思路是贪心。

题意：从一系列区间[a_i,b_i]中至少取出c_i个数构成集合s，求s的最小size？

对于每个区间，有前面已经选了的，还有自己要选的，对于自己要选的。尽量让自己选的靠后，就可以让后面的区间多选。所以按照末尾点排序，对于区间求和，用bit

#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define l(x) ((x)<<1)
#define r(x) ((x)<<1|1)
#define lowbit(x) ((x)&(-(x)))
#define abs(x) ((x)>=0?(x):(-(x)))
#define ms(a,b) memset(a,b,sizeof(a))

using namespace std;

const int maxn = 51111;

struct node {
	int a, b, val;
	friend bool operator<(const node &x, const node &y) {
		return x.b < y.b;
	}
};

int bit[maxn * 2 + 1];
int sum(int i) {
	int x = 0;
	while (i>0) {
		x += bit[i];
		i -= lowbit(i);
	}
	return x;
}

int sum(int a, int b) {
	return sum(b) - sum(a - 1);
}

void add(int i, int x) {
	while (i<= maxn) {
		bit[i] += x;
		i += lowbit(i);
	}
}

int n, a, b, c;
node demo[maxn];
int uesd[maxn];

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> demo[i].a >> demo[i].b >> demo[i].val;
	}
	sort(demo, demo + n);
	ms(uesd, 0);
	int res = 0,last,now;
	for (int i = 0; i < n; i++) {
		last = sum(demo[i].a, demo[i].b);
		now = demo[i].b;
		while (last < demo[i].val) {
			if (uesd[now] == 0) {
				uesd[now] = 1;
				add(now, 1);
				res++;
				last++;
			}
			now--;
		}
	}
	cout << res << "\n";

	return 0;
}