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Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1599 Accepted Submission(s): 443

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. (

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

题意:dudu要在不同班级里面选择两名队友参加ACM，如果两名队友的IQ1+IQ2>k (k是dudu的IQ)，那么就可以选择这两名队友参赛。问总共有多少种选法。

题解:对于某个班级,假设第 j 个人的IQ是 v,那么对应这个人可以找到的队友数等于

在所有班级里面找到 IQ>k-v 的人 - 在本班找到 IQ > K-v 的人把这些全部加起来就好了。

用一个数组记录所有班级的人IQ，然后从小到大排序，二分查找大于等于 k-v 的第一个人的位置。这里可以使用 lower_bound 来进行二分查找。最后结果要除以二,因为算重了。

时间复杂度O(n*m*log(n*m))

lower_bound:函数lower_bound()在first和last中的前闭后开区间进行二分查找，返回大于或等于val的第一个元素位置。如果所有元素都小于val，则返回last的位置。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;const int N = 1005;const int M = 105;int v[M];int v1[N*M];int sum[N];struct Node{    int num;    int v[N];}node[N];int main(){    int tcase,n;    int k;    scanf("%d",&tcase);    while(tcase--){        scanf("%d%d",&n,&k);        int cnt = 1;        for(int i=1;i<=n;i++){            scanf("%d",&node[i].num);            sum[i] = sum[i-1]+node[i].num;            for(int j=1;j<=node[i].num;j++){                scanf("%d",&v[j]);                v1[cnt++] = v[j];            }            sort(v+1,v+node[i].num+1);            for(int j=1;j<=node[i].num;j++){                node[i].v[j] = v[j];            }        }        sort(v1+1,v1+cnt);        long long ans = 0;        for(int i=1;i<=n;i++){            for(int j=1;j<=node[i].num;j++){                int v = node[i].v[j];                int pos1 = node[i].v+node[i].num+1 - lower_bound(node[i].v+1,node[i].v+1+node[i].num,k-v+1);                int pos2 = v1+cnt - lower_bound(v1+1,v1+cnt,k-v+1);                ans += pos2-pos1;            }        }        printf("%lld\n",ans/2);    }    return 0;}

转载于:https://www.cnblogs.com/liyinggang/p/5659607.html

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