hdu 5101 思维题

http://acm.hdu.edu.cn/showproblem.php?pid=5101

Problem Description

One day, Dudu, the most clever boy, heard of ACM/ICPC, which is a very interesting game. He wants to take part in the game. But as we all know, you can't get good result without teammates.
So, he needs to select two classmates as his teammates. 
In this game, the IQ is very important, if you have low IQ you will WanTuo. Dudu's IQ is a given number k. We use an integer v[i] to represent the IQ of the ith classmate. 
The sum of new two teammates' IQ must more than Dudu's IQ.
For some reason, Dudu don't want the two teammates comes from the same class.
Now, give you the status of classes, can you tell Dudu how many ways there are.

 

Input

There is a number T shows there are T test cases below. ( T≤20 )
For each test case , the first line contains two integers, n and k, which means the number of class and the IQ of Dudu. n (  0≤n≤1000  ), k(  0≤k<231  ).
Then, there are n classes below, for each class, the first line contains an integer m, which means the number of the classmates in this class, and for next m lines, each line contains an integer v[i], which means there is a person whose iq is v[i] in this class. m(  0≤m≤100  ), v[i](  0≤v[i]<231  )

 

Output

For each test case, output a single integer.

 

Sample Input

   
   
   
   

    
    
    
    1
3 1
1 2
1 2
2 1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    5
   
   
   
   

/**
HDU 5101
题目大意：
          给定一些集合，选择两个来自不同集合的数，加和大于k，问有多少种选择方案。
解题思路：
         答案=从所有数中选择的两个加和大于k的数的方案数-在同一个集合中选择的两个加和大于k的数的方案数
         而对于同一个集合中选择的两个加和大于k的方案数是可以直接排序然后利用单调性快速统计出来的。
*/
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long LL;
LL a[1005][105],sum[100005],v[1005];
LL n,m,cnt;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int k=0;
        scanf("%I64d%I64d",&n,&cnt);
        for(int i=0; i<n; i++)
        {
            scanf("%I64d",&v[i]);
            for(int j=0; j<v[i]; j++)
            {
                scanf("%I64d",&a[i][j]);
                sum[k++]=a[i][j];
            }
            sort(a[i],a[i]+v[i]);
        }
        sort(sum,sum+k);
        int ans=0;
        LL p=k-1;
        for(int i=0; i<k; i++)//每次找i位置以后和它和>cnt的数的个数，这样扫一遍保证不会有重复的
        {
            while(sum[i]+sum[p]>cnt&&p>i)
                p--;
            if(p>i)
                ans+=k-1-p;
            else
                ans+=k-1-i;
        }
        for(int i=0; i<n; i++)
        {
            int p=v[i]-1;
            for(int j=0; j<v[i]; j++)
            {

                while(sum[j]+sum[p]>cnt&&p>j)
                    p--;
                if(p>j)
                    ans-=v[i]-1-p;
                else
                    ans-=v[i]-1-j;
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}