Codeforces 161 D Distance in Tree 树形DP

题目大意：给出一棵n个节点的树，统计树中长度为k的路径的条数（1<=n<=50000
,1<=k<=500）；

题目分析：设dp[i][j]表示以i作为长度为j的路径其中一个点的方案数。转移：dp[u][j]+=dp[v][j-1];  方案计数：ans+=dp[v][j]*dp[u][k-j-1];

#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[100005];

int tot,first[100005],dp[50005][505],k,ans,n,x,y;

void add(int u,int v){
tot++;
e[tot].to=v;
e[tot].next=first[u];
first[u]=tot;
}

void dfs(int u,int fa){
dp[u][0]=1;

for (int p=first[u];p;p=e[p].next){
int v=e[p].to;
if (v==fa) continue;
dfs(v,u);

Rep (j,k) ans+=dp[v][j]*dp[u][k-j-1];
For (j,k) dp[u][j]+=dp[v][j-1];

}

}

int main(){
scanf("%d%d",&n,&k);
For (i,n-1){
scanf("%d%d",&x,&y);
add(x,y);
add(y,x);
}
dfs(1,0);
printf("%d",ans);
}