【bzoj3522】[Poi2014]Hotel 暴力+计数

来源：互联网发布：into you mac miller 编辑：程序博客网时间：2024/06/01 23:44

题目大意：给定一棵树，选取树上三个点，使得三个点任意两点之间距离相等，问方案数。

题目分析：对于任意三个点，要么处于一条链，或者不处于同一条链。显然处于一条链时不满足题意。所以必然这三个点是有一个中心使得该点到三个点距离相等。所以暴力枚举中心，再计算方案数。

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#include<cmath>#include<cctype>#include<cassert>#include<climits>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForD(i,n) for(int i=n;i;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define RepD(i,n) for(int i=n;i>=0;i--)#define MEM(a) memset(a,0,sizeof(a))#define MEMI(a) memset(a,127,sizeof(a))#define MEMi(a) memset(a,128,sizeof(a))#define INF (2139062143)#define phiF (1000000006)#define MAXN (1000000+10)typedef long long LL;struct info{int to,next;}e[10005];int first[5005],deep[5005],x,y,n,tot;LL ans,t1[5005],t2[5005];void add(int x,int y){e[++tot].to=y;e[tot].next=first[x];first[x]=tot;}void dfs(int u,int fa,int dep){deep[dep]++;for (int p=first[u];p;p=e[p].next){int v=e[p].to;if (fa!=v)dfs(v,u,dep+1);}}int main(){scanf("%d",&n);For (i,n-1){scanf("%d%d",&x,&y);add(x,y);add(y,x);}For (u,n){MEM(t1);MEM(t2);for (int p=first[u];p;p=e[p].next){MEM(deep);dfs(e[p].to,u,1);For (j,n){ans+=(LL)deep[j]*t2[j];t2[j]+=t1[j]*deep[j];t1[j]+=deep[j];}}}printf("%lld",ans);}

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