HDU6180-Schedule

Schedule

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 572 Accepted Submission(s): 219

Problem Description
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

Source
2017 Multi-University Training Contest - Team 10

题目大意：有一些任务，重叠的任务不能用同一台机器解决，每台机器只能开一次，问最少机器为多少？在最少机器的前提下，最少时间为多少？
解题思路： 用multiset保存结束时间，以现在任务的开始时间进行二分查找，若可以用以前的机器，则扩展那个结束时间，否则新开一台机器。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <map>#include <algorithm>#include <set>#define mem(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long LL;typedef pair<int,int> PII;const int MAXN=1e5+50;multiset<int> ms;struct node{    int st,ed;}p[MAXN];bool cmp(node x,node y){    if(x.st==y.st) return x.ed<y.ed;    return x.st<y.st;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        ms.clear();        int n; scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d%d",&p[i].st,&p[i].ed);        }        sort(p+1,p+n+1,cmp);//        for(int i=1;i<=n;i++)//        {//            cout<<p[i].st<<" "<<p[i].ed<<endl;//        }        LL ans=0;        ms.insert(p[1].ed);        ans+=p[1].ed-p[1].st;        for(int i=2;i<=n;i++)        {            auto pos=ms.upper_bound(p[i].st);            if(pos==ms.begin())            {                ms.insert(p[i].ed);                ans+=p[i].ed-p[i].st;            }else            {                auto q=pos;                q--;                ans+=p[i].ed-(*q);                ms.erase(q);                ms.insert(p[i].ed);            }        }        printf("%d %lld\n",ms.size(),ans);    }    return 0;}/*131 34 62 5*/