HDU6180(贪心)

Schedule

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 181    Accepted Submission(s): 73

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

 

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解题思路：首先把所有任务按照左端点排个序，然后从左往右扫，我们每次看当前任务的s值比我们已经用过的机器中t值是否要大，如果是，则选择一个机器的t值离s值最近的机器来完成这个任务，否则开一台新的机器。我们用一个multiset维护一下就行

#include <bits/stdc++.h>using namespace std;const int maxn = 100000 + 10;typedef long long LL;struct task{    LL s, t;    bool operator <(const task &res) const {        return s < res.s;    }}T[maxn];struct machine{    LL s, t;    bool operator <(const machine &res) const{        return t > res.t;    }};int n;//priority_queue<machine> q;multiset<machine> ss;multiset<machine>::iterator it;//machine term[maxn];int cnt;int main(){    //freopen("C:\\Users\\creator\\Desktop\\in1.txt","r",stdin) ;    //IO::Begin() ;    //freopen("C:\\Users\\creator\\Desktop\\out.txt","w",stdout) ;    int Case;    scanf("%d", &Case);    while(Case--)    {        ss.clear();        scanf("%d", &n);        for(int i = 1; i <= n; i++)        {            scanf("%lld%lld", &T[i].s, &T[i].t);        }        sort(T + 1, T + n + 1);        //while(!q.empty()) q.pop();        LL sum = 0;        int ans = 0;        machine test;        task temp;        for(int i = 1; i <= n; i++)        {            if(ss.empty())            {                ans++;                test.s = T[i].s;                test.t = T[i].t;                ss.insert(test);            }            else            {                test.s = 0;                test.t = T[i].s;                it = ss.lower_bound(test);                //it = lower_bound(ss.begin(), ss.end(), test);                if(it == ss.end())                {                    test.s = T[i].s;                    test.t = T[i].t;                    ss.insert(test);                    ans++;                }                else                {                    test = *it;                    test.t = T[i].t;                    ss.erase(it);                    ss.insert(test);                }            }        }        for(it = ss.begin(); it != ss.end(); ++it)        {            test = *it;            sum += test.t - test.s;        }        printf("%d %lld\n", ans, sum);    }    return 0;}