http://codeforces.com/contest/489/problem/D
很顯然，我們只需要找對于每個點能到達的深度為3的點的路徑的數量，那么對于一個深度為3的點，如果有a種方式到達，那么有方案數(a-1+1)*(a-1)/2
可是我用dfs找路徑就tle了QAQ
于是orz別人的代碼，，，，是暴力。。。。。。。。。。。。。。。。。。。。。。。。直接兩重循環orz
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
const int N=3005;
struct dat { int to, next; }e[N*10];
int cnt, vis[N], c[N], n, m, ihead[N];
void add(int u, int v) { e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; }
void bfs(int x, int dep) {
rdm(x, i) {
int y=e[i].to;
rdm(y, j) {
int z=e[j].to;
if(x==z) continue;
++c[z];
}
}
}
ll ans;
int main() {
read(n); read(m);
for1(i, 1, m) { int u=getint(), v=getint(); add(u, v); }
for1(i, 1, n) {
for1(j, 1, n) vis[j]=0, c[j]=0;
bfs(i, 1);
//for1(j, 1, n) cout << c[j] << ' '; cout << endl;
for1(j, 1, n) if(c[j]>=2) {
--c[j];
ans+=(ll)(c[j]+1)*c[j]/2;
}
}
printf("%I64d\n", ans);
return 0;
}
?
?
?
Tomash keeps wandering off and getting lost while he is walking along the streets of Berland. It's no surprise! In his home town, for any pair of intersections there is exactly one way to walk from one intersection to the other one. The capital of Berland is very different!
Tomash has noticed that even simple cases of ambiguity confuse him. So, when he sees a group of four distinct intersections?a,?b,?c?and?d, such that there are two paths from?a?to?c?— one through?b?and the other one through?d, he calls the group a "damn rhombus". Note that pairs?(a,?b),?(b,?c),?(a,?d),?(d,?c)?should be directly connected by the roads. Schematically, a damn rhombus is shown on the figure below:
Other roads between any of the intersections don't make the rhombus any more appealing to Tomash, so the four intersections remain a "damn rhombus" for him.
Given that the capital of Berland has?n?intersections and?m?roads and all roads are unidirectional and are known in advance, find the number of "damn rhombi" in the city.
When rhombi are compared, the order of intersections?b?and?d?doesn't matter.
The first line of the input contains a pair of integers?n,?m?(1?≤?n?≤?3000,?0?≤?m?≤?30000) — the number of intersections and roads, respectively. Next?m?lines list the roads, one per line. Each of the roads is given by a pair of integers?ai,?bi?(1?≤?ai,?bi?≤?n;ai?≠?bi) — the number of the intersection it goes out from and the number of the intersection it leads to. Between a pair of intersections there is at most one road in each of the two directions.
It is not guaranteed that you can get from any intersection to any other one.
Print the required number of "damn rhombi".
5 4
1 2
2 3
1 4
4 3
1
4 12
1 2
1 3
1 4
2 1
2 3
2 4
3 1
3 2
3 4
4 1
4 2
4 3
12
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