1307 - Counting Triangles
| 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 |
PROBLEM SETTER: JANE ALAM JAN
题意：给你的边能构成多少个三角形。
思路：先暴力组合两条边，然后二分查询第三条边即可。
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<queue> 6 #include<stdlib.h> 7 #include<math.h> 8 #include<stack> 9 #include<vector> 10 #include<map> 11 using namespace std; 12 typedef long long LL; 13 int ans[2005]; 14 typedef struct pp 15 { 16 short int x; 17 short int y; 18 } ss; 19 ss ak[4000005]; 20 int main(void) 21 { 22 int i,j,k; 23 scanf("%d",&k); 24 int s; 25 for(s=1; s<=k; s++) 26 { 27 int n,m; 28 scanf("%d",&n); 29 for(i=0; i<n; i++) 30 { 31 scanf("%d",&ans[i]); 32 } 33 sort(ans,ans+n); 34 int cnt=0; 35 for(i=0; i<n; i++) 36 { 37 for(j=i+1; j<n; j++) 38 { 39 ak[cnt].x=i; 40 ak[cnt].y=j; 41 cnt++; 42 } 43 } 44 LL sum=0; 45 for(i=0; i<cnt; i++) 46 { 47 int l=0; 48 int r=n-1; 49 int maxx=max(ans[ak[i].x],ans[ak[i].y]); 50 int minn=min(ans[ak[i].x],ans[ak[i].y]); 51 int id=0; 52 l=0; 53 r=n-1;int id1=-1; 54 while(l<=r) 55 { 56 int mid=(l+r)/2; 57 if(ans[mid]+minn>maxx) 58 { 59 id1=mid; 60 r=mid-1; 61 } 62 else l=mid+1; 63 } 64 l=0; 65 r=n-1;int id2=-1; 66 while(l<=r) 67 { 68 int mid=(l+r)/2; 69 if(ans[mid]<maxx+minn) 70 { 71 id2=mid; 72 l=mid+1; 73 } 74 else r=mid-1; 75 } 76 if(id1!=id2) 77 { 78 sum+=id2-id1+1; 79 if(ak[i].x>=id1&&ak[i].x<=id2) 80 sum--; 81 if(ak[i].y<=id2&&ak[i].y>=id1) 82 sum--; 83 } 84 } 85 printf("Case %d: ",s); 86 printf("%lld ",sum/3); 87 } 88 return 0; 89 }