ZOJ 3981 && 2017CCPC秦皇岛 A：Balloon Robot（思维题）

A - Balloon Robot

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3981

Description

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

BaoBao, an enthusiast for competitive programming, has made \(p\) predictions of the contest result before the contest. Each prediction is in the form of \((a_i,b_i)\), which means the \(a_i\)-th team solves a problem during the \(b_i\)-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 \(t_a\)-th time unit, and the balloon is sent to them during the \(t_b\)-th time unit, then the unhappiness of the team will increase by \(t_b-t_a\). 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 \(k\)-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 \(i\)-th (\(1 \le i < m\)) seat, it will move to the (\(i+1\))-th seat; If the robot is currently on the \(m\)-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 \(k\) 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 \(T\), indicating the number of test cases. For each test case:

The first line contains three integers \(n\), \(m\) and \(p\) (\(1 \le n \le 10^5\), \(n \le m \le 10^9\), \(1 \le p \le 10^5\)), indicating the number of participating teams, the number of seats and the number of predictions.

The second line contains \(n\) integers \(s_1, s_2, \dots, s_n\) (\(1 \le s_i \le m\), and \(s_i \ne s_j\) for all \(i \ne j\)), indicating the seat number of each team.

The following \(p\) lines each contains two integers \(a_i\) and \(b_i\) (\(1 \le a_i \le n\), \(1 \le b_i \le 10^9\)), indicating that the \(a_i\)-th team solves a problem at time \(b_i\) according to BaoBao's predictions.

It is guaranteed that neither the sum of \(n\) nor the sum of \(p\) over all test cases will exceed \(5 \times 10^5\).

Output

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

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

Sample Output

1
4
5
50

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：
第一行三个数字n, m, q表示有m个座位围成一个环，n个队伍，q次A题
接下来n个数表示n个队伍所在位置(1<=ai<=m)
再接下来q行，每行a, b表示第a个队伍在第b秒A了一道题
有一个只会每一秒顺时针移动一个位置的发气球机器人
只要当前队伍有题目已经A了就会给他对应数量的气球（当然每道题最多1个气球）
如果a队伍在b时刻A了一道题，并在c时刻才拿到气球，那么这个队伍就会积累c-b点不开心值
求一个机器人起始位置（一开始是第0秒）使得所有队伍最终不开心值之和最小

分析：
假设机器人就在位置1，可以O(n)求出所有人的不开心值，排个序
之后暴力枚举初始位置，每移动1个位置可以使得所有不开心值不为0的队伍不开心值-1，
不开心值为0的队伍不开心值变为m，因为排过序所以这个可以O(1)转移
复杂度O(m)
m太大但其实有些位置一定不可能是最优的，
所以理论上只用枚举最多q个位置即可

可以写一下机器从1 2 3 开始 各个气球点的等待时代，发现是每次-1 -1，0就变为m。

那么就假设从1开始，得到每个气球的等待数组b。
给b排个序，用重复的只要算一次。 让i这个点等于0
（即减了b[i]），那么i前面的点都加了m。
所有的点都减了b[i]。就可以遍历一次答案，取最小。

 

#include<stdio.h>
#include<iostream>
#include<vector>
#include <cstring>
#include <stack>
#include <cstdio>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include<string>
#include<math.h>
#define max_v 1000005
#define INF 999999999
using namespace std;
typedef long long LL;
LL a[max_v];
LL b[max_v];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        //n个队伍 m个座位 k次ac事件
        LL n,m,k;
        scanf("%lld %lld %lld",&n,&m,&k);

        //n个队伍的位置
        for(LL i=;i<=n;i++)
            scanf("%lld",&a[i]);

        LL id,time;//队伍id和ac某题的时间
        LL sum=;
        for(LL i=;i<=k;i++)//假设机器人位置在1 得到所有队伍的不高兴值排序
        {
            scanf("%lld %lld",&id,&time);
            time=time%m;
            if(time==)
                time=m;
            b[i]=(a[id]-time+m)%m;
            sum+=b[i];
        }

        sort(b+,b+k+);//将所有队伍的不高兴值排序
        long long  ans=0x3f3f3f3f3f3f3f3f;
        b[]=-;
        for(LL i=;i<=k;i++)
        {
            if(b[i]!=b[i-])
                ans=min(ans,sum-k*b[i]+(i-)*m);
        }
        printf("%lld\n",ans);
    }
    return ;
}
/*
题意：
第一行三个数字n, m, q表示有m个座位围成一个环，n个队伍，q次A题
接下来n个数表示n个队伍所在位置(1<=ai<=m)
再接下来q行，每行a, b表示第a个队伍在第b秒A了一道题
有一个只会每一秒顺时针移动一个位置的发气球机器人
只要当前队伍有题目已经A了就会给他对应数量的气球（当然每道题最多1个气球）
如果a队伍在b时刻A了一道题，并在c时刻才拿到气球，那么这个队伍就会积累c-b点不开心值
求一个机器人起始位置（一开始是第0秒）使得所有队伍最终不开心值之和最小

分析：
假设机器人就在位置1，可以O(n)求出所有人的不开心值，排个序
之后暴力枚举初始位置，每移动1个位置可以使得所有不开心值不为0的队伍不开心值-1，
不开心值为0的队伍不开心值变为m，因为排过序所以这个可以O(1)转移
复杂度O(m)
m太大但其实有些位置一定不可能是最优的，
所以理论上只用枚举最多q个位置即可

可以写一下机器从1 2 3 开始 各个气球点的等待时代，发现是每次-1 -1，0就变为m。

那么就假设从1开始，得到每个气球的等待数组b。
给b排个序，用重复的只要算一次。 让i这个点等于0
（即减了b[i]），那么i前面的点都加了m。
所有的点都减了b[i]。就可以遍历一次答案，取最小。

*/