HDU-5101-Select（二分）

题目连接：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

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题目大意，给出T组测试数据，n个班级和一个人的智商k，对于每个班级，首先输入这个班的人数，然后按顺序输入这些人的智商，要找两个分别在不同班级的人&&他们的智商之和大于k，问有多少中组合。

分析，如果对于每个人都找一遍的话，用二分，复杂度为O（n*m*log（n*m）），看看数据范围，可行

使用二分总结中的 5.查找第一个大于key的元素的位置和 6.查找最后一个小于等于key的元素的位置

//#pragma comment(linker, "/STACK:1024000000,1024000000") 
 
#include<stdio.h>
#include<string.h>  
#include<math.h>  
  
#include<map>   
//#include<set>
#include<deque>  
#include<queue>  
#include<stack>  
#include<bitset> 
#include<string>  
#include<fstream>
#include<iostream>  
#include<algorithm>  
using namespace std;  
 
#define ll long long  
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b) 
#define clean(a,b) memset(a,b,sizeof(a))// 水印 
//std::ios::sync_with_stdio(false);
const int MAXN=1e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);

ll room[1010][110],arr[MAXN];
ll n,k;

int main()
{
	//std::ios::sync_with_stdio(false);
	int T;
	while(cin>>T)
	{
		while(T--)
		{
			clean(arr,0);
			clean(room,0);
			scanf("%lld%lld",&n,&k);
			int tot=0;
			for(int i=1;i<=n;++i)
			{
				scanf("%lld",&room[i][0]);
				for(int j=1;j<=room[i][0];++j)
				{
					scanf("%lld",&room[i][j]);
					//cin>>room[i][j];
					arr[tot++]=room[i][j];
				}
				sort(room[i]+1,room[i]+room[i][0]+1);
			}
			sort(arr,arr+tot);//默认从低到高 
			ll ans=0;
//			for(int i=1;i<=n;++i)
//				cout<<room[i][0]<<endl;
			for(int i=1;i<=n;++i)
			{
				//对于每个人，找出第一个大于k的位置 
				//再减去本班符合要求的人数 
				for(int j=1;j<=room[i][0];++j)
				{
					int l=0,r=tot-1,mid;
					while(l<=r)
					{
						mid=(l+r)>>1;
						if(room[i][j]+arr[mid]>k)
							r=mid-1;
						else if(room[i][j]+arr[mid]<=k)
							l=mid+1;
					}
					//l为第一个大于k的位置 
					ll res1=tot-l;
					//找出最后一个小于等于k的位置 
					l=1,r=room[i][0];
					while(l<=r)
					{
						mid=(l+r)>>1;
						if(room[i][j]+room[i][mid]>k)
							r=mid-1;
						else if(room[i][j]+room[i][mid]<=k)
							l=mid+1;
					}
					//r为最后一个位置 
					ll res2=room[i][0]-r;
					//cout<<res1<<" "<<res2<<endl;
					ans+=(res1-res2);
				}
			}
			printf("%lld\n",ans/2);
			//cout<<ans/2<<endl;
		}
	}
	
	
}

/*
Sample Input

1
3 1
1 2
1 2
2 1 1

Sample Output

5



*/

HDU-5101-Select（二分）

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