HDU 6180 Schedule（贪心）

作者：CSDN博主 最菜的acmer

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原题

Schedule

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 153428/153428 K (Java/Others)

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

题意

给出任务的开始时间和结束时间，一台机器同一时间只能做一个任务，且只能开关机一次，问最少需要多少台机器和对应的最少的工作时间之和。

涉及知识及算法

按任务开始时间从小到大排序后，从头开始加入set容器：如果任务开始时间大于等于set里的最近的结束时间，就插入 ，否则就重新开一台机器即可。

代码

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<vector>#include<algorithm>#include<string>#include <set>using namespace std;typedef long long LL;const int MAX_N = 1e5+5;struct node{    int s,e;};node a[MAX_N];bool cmp(node a,node b){    return a.s<b.s;}multiset<int>st;int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        st.clear();        for(int i=0;i<n;i++)        {            scanf("%d%d",&a[i].s,&a[i].e);        }        sort(a,a+n,cmp);        LL ans=0;        for(int i=0;i<n;i++)        {            multiset<int>::iterator it=st.upper_bound(a[i].s);            //如果没找到            if(it==st.begin())            {                //开一台机器                ans+=a[i].e-a[i].s;                st.insert(a[i].e);            }            //如果找到            else            {                //加入                it--;                ans+=a[i].e-*it;                st.erase(it);                st.insert(a[i].e);            }        }        printf("%d %lld\n",st.size(),ans);    }    return 0;}