【杭电oj】5101 - Select（STL & 二分）

最新推荐文章于 2019-07-22 22:57:49 发布

wyg1997

最新推荐文章于 2019-07-22 22:57:49 发布

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本文介绍了在杭电oj平台上解决5101题目的过程，初次尝试因剪枝不足导致超时，后采用从全部人选中寻找两两和大于k的策略，再排除班级内满足条件的，最终结果除以2，实现了通过二分查找优化的解决方案。

点击打开题目

Select

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1581 Accepted Submission(s): 441

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

Sample Output

Source

BestCoder Round #17

第一次在每个班找人，再去其他班找人，最后得结果，然后就 TLE 了，虽然做了个剪枝。

后来的思路是：在全部人中找两两之和大于 k 的，然后减去一个班里满足条件的，最后除以 2 就行了。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int iq[1011][111];
int stu[100000+11];
int main()
{
	int u;
	int n,k;
	int m;
	__int64 ans;
	int ant;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d %d",&n,&k);
		ans = m = 0;
		for (int i = 0 ; i < n ; i++)
		{
			scanf ("%d",&iq[i][0]);
			for (int j = 1 ; j <= iq[i][0] ; j++)
			{
				scanf ("%d",&iq[i][j]);
				stu[m++] = iq[i][j];
			}
			sort (iq[i] + 1 , iq[i] + iq[i][0] + 1); 
		}
		sort (stu , stu + m);
		for (int i = m - 1 ; i >= 0 ; i--)
		{
			ant = m - (upper_bound (stu , stu + m , k - stu[i]) - stu);
			if (ant)
				ans += ant;
			else
				break;
		}
		for (int i = 0 ; i < n ; i++)
		{
			for (int j = iq[i][0] ; j >= 1 ; j--)
			{
				ant = iq[i][0] - (upper_bound (iq[i] + 1 , iq[i] + 1 + iq[i][0] , k - iq[i][j]) - iq[i]) + 1;
				if (ant)
					ans -= ant;
				else
					break;
			}
		}
		ans >>= 1;		//计算的结果是答案的二倍 
		printf ("%I64d\n",ans);
	}
	return 0;
}