EZOJ #80

传送门

分析

经典的树型DP

我们记录dp[i][0/1]表示i的子树中到i的长度分别为偶数和奇数的长度和

dp2[i][0/1]则表示不在i的子树中的点到i的长度分别为偶数和奇数的长度和

然后根据边的长度情况转移一下就可以了

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
long long dp[][],dp2[][];
int sum[][],sum2[][];
int val[],to[],head[];
int nxt[],cnt,n,m;
inline void add(int x,int y,int z){
    nxt[++cnt]=head[x];
    head[x]=cnt;
    to[cnt]=y;
    val[cnt]=z;
    nxt[++cnt]=head[y];
    head[y]=cnt;
    to[cnt]=x;
    val[cnt]=z;
}
inline void dfs1(int x,int fa){
    int i,j,k;
    for(i=head[x];i;i=nxt[i]){
      if(to[i]==fa)continue;
      dfs1(to[i],x);
      if(val[i]%){
          dp[x][]+=dp[to[i]][];
          dp[x][]+=1ll*sum[to[i]][]*val[i];
          sum[x][]+=sum[to[i]][];
          dp[x][]+=dp[to[i]][]+1ll*val[i];
          dp[x][]+=1ll*sum[to[i]][]*val[i];
          sum[x][]+=sum[to[i]][]+;
      }else {
          dp[x][]+=dp[to[i]][]+1ll*val[i];
          dp[x][]+=1ll*sum[to[i]][]*val[i];
          sum[x][]+=sum[to[i]][]+;
          dp[x][]+=dp[to[i]][];
          dp[x][]+=1ll*sum[to[i]][]*val[i];
          sum[x][]+=sum[to[i]][];
      }
    }
    return;
}
inline void dfs2(int x,int fa){
    int i,j,k;
    long long t0=,t1=;
    int s0=,s1=;
    for(i=head[x];i;i=nxt[i]){
      if(to[i]==fa)continue;
      if(val[i]%){
          t0+=dp[to[i]][];
          t0+=1ll*sum[to[i]][]*val[i];
          s0+=sum[to[i]][];
          t1+=dp[to[i]][]+1ll*val[i];
          t1+=1ll*sum[to[i]][]*val[i];
          s1+=sum[to[i]][]+;
      }else {
          t0+=dp[to[i]][]+1ll*val[i];
          t0+=1ll*sum[to[i]][]*val[i];
          s0+=sum[to[i]][]+;
          t1+=dp[to[i]][];
          t1+=1ll*sum[to[i]][]*val[i];
          s1+=sum[to[i]][];
      }
    }
    for(i=head[x];i;i=nxt[i]){
      if(to[i]==fa)continue;
      long long T0=t0,T1=t1;int S0=s0,S1=s1;
      if(val[i]%){
          T0-=dp[to[i]][];
          T0-=1ll*sum[to[i]][]*val[i];
          S0-=sum[to[i]][];
          T1-=dp[to[i]][]+1ll*val[i];
          T1-=1ll*sum[to[i]][]*val[i];
          S1-=sum[to[i]][]+;
          dp2[to[i]][]+=dp2[x][]+T1;
          dp2[to[i]][]+=1ll*(sum2[x][]+S1)*val[i];
          sum2[to[i]][]+=sum2[x][]+S1;
          dp2[to[i]][]+=dp2[x][]+1ll*val[i]+T0;
          dp2[to[i]][]+=1ll*(sum2[x][]+S0)*val[i];
          sum2[to[i]][]+=sum2[x][]++S0;
          dfs2(to[i],x);
      }else {
          T0-=dp[to[i]][]+1ll*val[i];
          T0-=1ll*sum[to[i]][]*val[i];
          S0-=sum[to[i]][]+;
          T1-=dp[to[i]][];
          T1-=1ll*sum[to[i]][]*val[i];
          S1-=sum[to[i]][];
          dp2[to[i]][]+=dp2[x][]+1ll*val[i]+T0;
          dp2[to[i]][]+=1ll*(sum2[x][]+S0)*val[i];
          sum2[to[i]][]+=sum2[x][]++S0;
          dp2[to[i]][]+=dp2[x][]+T1;
          dp2[to[i]][]+=1ll*(sum2[x][]+S1)*val[i];
          sum2[to[i]][]+=sum2[x][]+S1;
          dfs2(to[i],x);
      }
    }
    return;
}
int main(){
    int i,j,k,x,y,z;
    scanf("%d%d",&n,&m);
    for(i=;i<n;i++){
      scanf("%d%d%d",&x,&y,&z);
      add(x,y,z);
    }
    dfs1(,);
    dfs2(,);
    for(i=;i<=m;i++){
      scanf("%d",&x);
      printf("%lld %lld\n",dp[x][]+dp2[x][],dp[x][]+dp2[x][]);
    }
    return ;
}