HDU 6180 Schedule (扫描线化线成点)-优快云博客

本文链接：https://blog.youkuaiyun.com/Mr_Treeeee/article/details/77531883

本文介绍了一种解决多个任务在有限数量的机器上进行最优调度的问题。通过扫描线算法找到最少的机器数量及其总运行时间，确保任务不重叠且机器利用率达到最高。

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Schedule

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 153428/153428 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

(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

Sample Output

2 8

题意：

给你几个机器，有N个从l到r的任务，让你求最小数量的机器，可以让任务不延期完成，再求机器运行的时间。

机器不能中途停止！

POINT：

其实就是给你几个线段求最大重复。

扫描线，化线成点，一条线化为l点和r点。给这些点排个序。开始从最低扫描。cnt记录需要的机器数。

遇l，cnt++，遇r,cnt--。注意l与r重合的情况（不同线段的l和r），这题是一个任务结束之后可以立马开始另一个任务。

所以我们优先考虑r。我自以为是不可以的，WA了3次。

然后就是要记录每个机器的最左和最右：

最左肯定是这个机器第一次做任务。最右就遇r点更新一下就好了。ans+=rr-ll。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<iostream>
#include<algorithm>
#define maxn 100100
#define maxm 100100
#define LL long long
using namespace std;
typedef pair<LL,LL> pr;
vector<pr>len;
int main()
{
    LL T;
    scanf("%lld",&T);
    while(T--)
    {
        len.clear();
        LL n;
        scanf("%lld",&n);
        LL tm=0;
        for(LL i=1;i<=n;i++)
        {
            LL l,r;scanf("%lld %lld",&l,&r);
            len.push_back(make_pair(l,1));
            len.push_back(make_pair(r,-1));
        }
        LL maxk=0;
        LL cnt=0;
        sort(len.begin(),len.end());
        LL ll[maxn],rr[maxn];
        memset(ll,0,sizeof ll);
        memset(rr,0,sizeof rr);
        for(LL i=0;i<len.size();i++)
        {
            if(len[i].second==1)
            {
                cnt++;
                if(maxk<cnt)
                {
                    if(ll[cnt]==0)
                        ll[cnt]=len[i].first;
                   maxk=cnt;
                }
            }
            else
            {
                rr[cnt]=len[i].first;
                if(maxk<cnt) maxk=cnt;
                cnt--;
            }
        }
        for(LL i=1;i<=maxk;i++)
        {
            tm+=(rr[i]-ll[i]);
        }
        printf("%lld %lld\n",maxk,tm);
    }
  
    
    return 0;
}