2017 CCPC秦皇岛 A题 A Ballon Robot

The 2017 China Collegiate Programming Contest Qinhuangdao Site is coming! There will be  teams participating in the contest, and the contest will be held on a huge round table with  seats numbered from 1 to  in clockwise order around it. The -th team will be seated on the -th seat.

BaoBao, an enthusiast for competitive programming, has made  predictions of the contest result before the contest. Each prediction is in the form of , which means the -th team solves a problem during the -th time unit.

As we know, when a team solves a problem, a balloon will be rewarded to that team. The participants will be unhappy if the balloons take almost centuries to come. If a team solves a problem during the -th time unit, and the balloon is sent to them during the -th time unit, then the unhappiness of the team will increase by . In order to give out balloons timely, the organizers of the contest have bought a balloon robot.

At the beginning of the contest (that is to say, at the beginning of the 1st time unit), the robot will be put on the -th seat and begin to move around the table. If the robot moves past a team which has won themselves some balloons after the robot's last visit, it will give all the balloons they deserve to the team. During each unit of time, the following events will happen in order:

1. The robot moves to the next seat. That is to say, if the robot is currently on the -th () seat, it will move to the ()-th seat; If the robot is currently on the -th seat, it will move to the 1st seat.
2. The participants solve some problems according to BaoBao's prediction.
3. The robot gives out balloons to the team seated on its current position if needed.

BaoBao is interested in minimizing the total unhappiness of all the teams. Your task is to select the starting position  of the robot and calculate the minimum total unhappiness of all the teams according to BaoBao's predictions.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains three integers ,  and  (, , ), indicating the number of participating teams, the number of seats and the number of predictions.

The second line contains  integers  (, and  for all ), indicating the seat number of each team.

The following  lines each contains two integers  and  (, ), indicating that the -th team solves a problem at time  according to BaoBao's predictions.

It is guaranteed that neither the sum of  nor the sum of  over all test cases will exceed .

<h4< dd="">Output

For each test case output one integer, indicating the minimum total unhappiness of all the teams according to BaoBao's predictions.

<h4< dd="">Sample Input

4
2 3 3
1 2
1 1
2 1
1 4
2 3 5
1 2
1 1
2 1
1 2
1 3
1 4
3 7 5
3 5 7
1 5
2 1
3 3
1 5
2 5
2 100 2
1 51
1 500
2 1000

<h4< dd="">Sample Output

1
4
5
50

<h4< dd="">Hint

For the first sample test case, if we choose the starting position to be the 1st seat, the total unhappiness will be (3-1) + (1-1) + (6-4) = 4. If we choose the 2nd seat, the total unhappiness will be (2-1) + (3-1) + (5-4) = 4. If we choose the 3rd seat, the total unhappiness will be (1-1) + (2-1) + (4-4) = 1. So the answer is 1.

For the second sample test case, if we choose the starting position to be the 1st seat, the total unhappiness will be (3-1) + (1-1) + (3-2) + (3-3) + (6-4) = 5. If we choose the 2nd seat, the total unhappiness will be (2-1) + (3-1) + (2-2) + (5-3) + (5-4) = 6. If we choose the 3rd seat, the total unhappiness will be (1-1) + (2-1) + (4-2) + (4-3) + (4-4) = 4. So the answer is 4.

题意：就是CCPC比赛，有n支队伍，m个座位,n支队伍坐在这些座位上面，一个机器人从任意位置开始每次移动一个位置分发气球，如果到了m，则下一次从新回到第一个位置；每只队伍有一个

愤怒值为过题时间和分发气球的时间差，让你求这些队伍愤怒值和的最小值；

题解：窝萌阔以假设它从1号店开始，算出每次请求的t-b的值并保存在数组h中，值的范围在（0~m-1）之间。若起点向后移动一个则数组h中的数据都加一，且等于M的都变为0。由于M有10的9次方所以不能遍历所有的可能。

所以将h数组排序，每次将最大值变为0，即整个数组都加上（m-最大值）

参考代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 typedef long long ll;
 const int maxn=1e5+;
 int t;
 ll n,m,p,x,y,a[maxn],h[maxn];
 int main()
 {
     cin>>t;
     while(t--)
     {
         ll ans=;
         cin>>n>>m>>p;
         for(int i=;i<=n;i++) cin>>a[i];
         for(int i=;i<p;i++)
         {
             cin>>x>>y;
             h[i]=(a[x]-(y%m)+m)%m;
             ans+=h[i];
         }
         ll sum=ans,cp=;
         sort(h,h+p);
         for(int i=p-;i>=;i--)
         {
             int temp=;
             while(h[i-temp]==h[i]&&i>=temp) temp++;
             h[i]+=cp;  cp+=m-h[i];
             ans+=(p-temp)*(m-h[i]); ans-=temp*h[i];
             sum=min(sum,ans);
             i-=temp-;
         }
         cout<<sum<<endl;
     }
     return ;
 }