light oj 1307 - Counting Triangles (二分--三角形个数)

1307 - Counting Triangles

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You are given N sticks having distinct lengths; you have to form some triangles using the sticks. A triangle is valid if its area is positive. Your task is to find the number of ways you can form a valid triangle using the sticks.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integer N (3 ≤ N ≤ 2000). The next line contains N integers denoting the lengths of the sticks. You can assume that the lengths are distinct and each length lies in the range[1,
109].

Output

For each case, print the case number and the total number of ways a valid triangle can be formed.

Sample Input	Output for Sample Input
3 5 3 12 5 4 9 6 1 2 3 4 5 6 4 100 211 212 121	Case 1: 3 Case 2: 7 Case 3: 4

 思路：O(n^3)必然超时，更少的时间就是O(n^2 logn)了，嗯，两层for循环，利用三角形边的性质 两边之和大于第三边（两边之差小于第三边），先排序，在二分查找

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
int T,n,kcase=1;
long long a[2010];
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
sort(a+1,a+n+1);
long long ans=0;
for(int i=1;i<=n-2;i++)
{
for(int j=i+1;j<=n-1;j++)
{
long long s=a[i]+a[j];
int pos=lower_bound(a+j+1,a+n+1,s)-a;
ans+=pos-j-1; //保留符合条件的值
}
}
printf("Case %d: %lld\n",kcase++,ans);
}
return 0;
}