HDU6180-Schedule

Schedule

                                                                     Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
                                                                                             Total Submission(s): 1042    Accepted Submission(s): 420

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

131 34 62 5

 

Sample Output

2 8

 

Source

2017 Multi-University Training Contest - Team 10

 

题意：有n个需要全部完成工作（时间区间），每台机器同时只能做一份工作，并且机器开机后，除非做完最后一个作业，不然不关机。求在最少机器数的情况下，最少的工作的总时间。

解题思路：首先按照开始时间排序，一个一个工作的扫过去，对于每一个工作，找到结束时间最晚的并且小于等于这个工作的开始时间，让这个机器做这个工作就可以了，二分一些就好了。Algorithm里面的二分比set自带的二分要慢很多

#include <iostream>  #include <cstdio>  #include <string>  #include <cstring>  #include <algorithm>  #include <queue>  #include <vector>  #include <set>  #include <stack>  #include <map>  #include <climits>  #include <functional>  using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;inline char get(){    static char buf[100000], *p1 = buf, *p2 = buf;    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;}template <class T> inline bool read(T & x){    char ch = get();    if (ch == EOF) return false;    while (ch<'0' || ch>'9') ch = get();    x = ch - '0';    while ((ch = get()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';    return true;}struct node{    int l, r;    bool operator<(const node &a)const    {        return l < a.l;    }}a[100005];multiset<int> s;int n;int main(){    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d", &n);        for (int i = 0; i < n; i++) scanf("%d%d", &a[i].l, &a[i].r);        sort(a, a + n);        LL ans = 0;        s.clear();        for (int i = 0; i < n; i++)        {            multiset<int>::iterator it = s.upper_bound(a[i].l);            if (it == s.begin())            {                ans += 1LL * (a[i].r - a[i].l);                s.insert(a[i].r);            }            else            {                it--;                ans += 1LL * (a[i].r - *it);                s.erase(it);                s.insert(a[i].r);            }        }        printf("%d %lld\n", s.size(), ans);    }    return 0;}