2017 Multi-University Training Contest 10 1010 Schedule && HDU 6180 (贪心)

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 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: 有若干个活动，第i个开始时间和结束时间是[Si,Ei) , 同一个教室安排的活动之间不能交叠，求要安排所有活动，

最少需要几个教室？ 并输出所有教室的使用时间之和，每个教室的使用时间是该教师最后一个活动结束时间-第一个

活动开始时间。

思路：对每个活动拆成两点，对所有点排序，On遍历一遍，碰到左端点cnt++，右端点cnt--，维护cnt的最大值即为所

需的教室数量，遍历同时计算时间和，这里要贪心一下，每个左端点都要找离他最近的右端点， 画几个图发现这样不会影响最后答案，用一个栈记录离他最近的右端点。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
struct node
{
ll x;
int flag;
bool operator < (const node &a) const
{
if(x == a.x) return flag > a.flag;
else return x < a.x;
}
}q[2*maxn];

int main(void)
{
int t, n;
cin >> t;
while(t--)
{
scanf("%d", &n);
ll ans = 0;
int num = 1;
for(int i=1;i<=n;i++)
{
scanf("%lld %lld",&q[num].x, &q[num+1].x);
q[num].flag = 1, q[num+1].flag = 2;
ans += q[num+1].x-q[num].x;
num += 2;
}
sort(q+1, q+num);
ll k = - 1, cnt = 0;
stack<ll>s;
while(!s.empty()) s.pop();
for(int i = 1; i < num; i++)
{
if(q[i].flag == 1)
{
cnt++;
if(!s.empty())
{
ll tmp = s.top(); s.pop();
ans += q[i].x - tmp;
}
}
else
{
cnt--;
s.push(q[i].x);
}
k = max(cnt, k);
}
printf("%lld %lld\n", k, ans);
}
return 0;
}