HDU 5100 Select 排序相加

Select

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 565    Accepted Submission(s): 167

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

13 11 21 22 1 1

 

Sample Output

5

/*HDU 5101  看时间长 刚开始暴力解决的 被人hack了。。。 int可以表示2^31啊 为什么这里非要用LL才能过？谁看了跟我说下 对每组进行排序，然后可以在O（n）时间求出同组相加大于k的数量。然后对整体排序，用O（N）求出所有的两两相加大于k的数量，然后相减。 */#include<iostream>#include<algorithm>#include<string>#include<stdio.h>#include<stack>#include<vector>typedef __int64 LL;using namespace std;LL v[1010][111];LL sum[1005*105];int cnt[1005];bool cmp(int a,int b)  {          return a<b;  }  int main(){    int t,i,j,tot,n,ii,jj;LL k,del,ans; //freopen("test.txt","r",stdin);    scanf("%d",&t);    while(t--)    {    scanf("%d%I64d",&n,&k);    tot=0;    del=0;    ans=0;    for(i=0;i<n;i++)    {    scanf("%d",&cnt[i]);    for(j=0;j<cnt[i];j++)    {    scanf("%I64d",&v[i][j]);    sum[tot++]=v[i][j];    }    sort(v[i],v[i]+cnt[i],cmp);    LL *vv=v[i];    for(ii=0,jj=cnt[i]-1;ii<jj;)//每个     {    if(vv[ii]+vv[jj]>k)    {    del+=(jj-ii);//说明j与i后面每个都可以大于jj--;     }    else    ii++;    }    }    sort(sum,sum+tot,cmp);    for(i=0,j=tot-1;i<j;)//总的     {    if(sum[i]+sum[j]>k)    {    ans+=(j-i);j--;     }   else    i++;    }ans-=del;printf("%I64d\n",ans);    }    return 0;}