bzoj 3522: [Poi2014]Hotel

呵呵，一开始天真的我以为求个 西格玛 C（？？，3）就好了。。

（题解：比枚举2个数的再多一个，，一样搞）

#include <bits/stdc++.h>
#define LL long long
#define lowbit(x) x&(-x)
#define inf 1e15
using namespace std;
inline int ra()
{
    int x=0,f=1; char ch=getchar();
    while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
struct edge{int next,to;}e[10005];
int tot[5005],mx,n,head[5005],cnt,s1[5005],s2[5005];
LL ans;
void insert(int x, int y){e[++cnt].next=head[x]; e[cnt].to=y; head[x]=cnt;}
void dfs(int x, int fa, int deep)
{
    tot[deep]++;
    mx=max(mx,deep);
    for (int i=head[x];i;i=e[i].next)
    {
        if (e[i].to==fa) continue;
        dfs(e[i].to,x,deep+1);
    }
}
int main(int argc, char const *argv[])
{
    n=ra();
    for (int i=1; i<n; i++)
    {
        int x=ra(),y=ra();
        insert(x,y); insert(y,x);
    }
    for (int x=1; x<=n; x++)
    {
        memset(s1,0,sizeof(s1));
        memset(s2,0,sizeof(s2));
        for (int i=head[x];i;i=e[i].next)
        {
            dfs(e[i].to,x,1);
            for (int j=1; j<=mx; j++)
            {
                ans+=s2[j]*tot[j];
                s2[j]+=s1[j]*tot[j];
                s1[j]+=tot[j];
            }
            for (int j=1; j<=mx; j++) tot[j]=0;
        }
    }
    cout<<ans;
    return 0;
}