数学专项counting:LA 5846
2013-03-26 22:48
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lrj书上已经给出了非常详细的讨论,结果即为C(n,3)-1/2*Sum(ai*(n-1-ai)),其中ai指的是每个点所连红边(或蓝边都行)的数量。
#include <iostream> #include <fstream> #include <cstring> #include <cstdio> using namespace std; #define M 1010 int n; int a[M]; int main() { freopen("in.txt","r",stdin); int T; cin>>T; while(T--) { memset(a,0,sizeof(a)); cin>>n; int c; for(int i=n;i>1;i--) { for(int j=1;j<i;j++) { cin>>c; if(c) { a[n-i+1]++; a[n-i+j+1]++; } } } int ans=0; for(int i=1;i<=n;i++) ans+=a[i]*(n-1-a[i]); ans=n*(n-1)*(n-2)/6-ans/2; cout<<ans<<endl; } return 0; }
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