Schedule HDU - 6180 (multiset ， 贪心)

There are N schedules, the i-th schedule has start time si and end time ei (1 <= i <= N). There are some machines. Each two overlapping schedules cannot be performed in the same machine. For each machine the working time is defined as the difference between timeend and timestart , where time_{end} is time to turn off the machine and timestart is time to turn on the machine. We assume that the machine cannot be turned off between the timestart and the timeend.
Print the minimum number K of the machines for performing all schedules, and when only uses K machines, print the minimum sum of all working times.
Input
The first line contains an integer T (1 <= T <= 100), the number of test cases. Each case begins with a line containing one integer N (0 < N <= 100000). Each of the next N lines contains two integers si and ei (0<=si<ei<=1e9).
Output
For each test case, print the minimum possible number of machines and the minimum sum of all working times.
Sample Input
1
3
1 3
4 6
2 5
Sample Output
2 8

题意：

思路：

先把每一个任务按照起始时间排序，然后用一个multiset维护每一个机器在做任务的终止时间，
对于每一个任务，我们去multiset中去二分找是否有一个机器的任务终止时间在当前任务开始时间之前（或等于）。
如果有，就在那个机器中执行该任务，否则就创建一个新机器。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int t;
struct node
{
	int l, r;

} a[maxn];
bool cmp(node &aa, node &bb)
{
	// if(aa.r!=bb.r)
	//     return aa.r<bb.r;
	// else
	return aa.l < bb.l;
}
int n;
multiset<int> st;
ll ans = 0ll;
int main()
{
	//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
	//freopen("D:\\common_text\code_stream\\out.txt","w",stdout);

	gg(t);
	while (t--)
	{
		gg(n);
		repd(i, 1, n)
		{
			gg(a[i].l);
			gg(a[i].r);
		}
		sort(a + 1, a + 1 + n, cmp);
		st.clear();
		ans = a[1].r - a[1].l;
		st.insert(a[1].r);
		repd(i, 2, n)
		{
			auto it = st.upper_bound(a[i].l);
			if (it == st.begin())
			{
				st.insert(a[i].r);
				ans += a[i].r - a[i].l;

			} else
			{
				it--;

				ans += a[i].r - (*it);
				st.erase(it);
				st.insert(a[i].r);

			}
		}
		cout << sz(st) << " " << ans << endl;
	}


	return 0;
}

inline void getInt(int* p) {
	char ch;
	do {
		ch = getchar();
	} while (ch == ' ' || ch == '\n');
	if (ch == '-') {
		*p = -(getchar() - '0');
		while ((ch = getchar()) >= '0' && ch <= '9') {
			*p = *p * 10 - ch + '0';
		}
	}
	else {
		*p = ch - '0';
		while ((ch = getchar()) >= '0' && ch <= '9') {
			*p = *p * 10 + ch - '0';
		}
	}
}