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HDU 5101 Select

2024-08-01 09:51:39   219人阅读

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Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 786    Accepted Submission(s): 232


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
 

Source
BestCoder Round #17
 

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。
//546MS	1860K
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
long long s[1007][107],num[1007],array[100007];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        long long k,a=0,ans=0;
        scanf("%d%I64d",&n,&k);
        for(int i=0; i<n; i++)
        {
            scanf("%I64d",&num[i]);
            for(int j=0; j<num[i]; j++)
            {
                scanf("%I64d",&s[i][j]);
                array[a++]=s[i][j];
            }
            sort(s[i],s[i]+num[i]);
        }
        sort(array,array+a);
        int x=a-1;
        for(int i=0; i<a; i++)//从所有数中选择的两个加和大于k的数的方案数
        {
            while(array[i]+array[x]>k&&x>i)x--;
            if(i>x)ans+=(a-i-1);
            else ans+=(a-x-1);
        }
        long long count=0;
        for(int i=0; i<n; i++)//在同一个集合中选择的两个加和大于k的数的方案数
        {
            int x=num[i]-1;
            for(int j=0; j<num[i]; j++)
            {
                while(s[i][j]+s[i][x]>k&&x>j) x--;
                if(j>=x)count+=(num[i]-j-1);
                else count+=(num[i]-x-1);
            }
        }
        printf("%I64d\n",ans-count);
    }
}


HDU 5101 Select


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