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[100100][2],dp2[100100][2];
int sum[100100][2],sum2[100100][2];
int val[200200],to[200200],head[200200];
int nxt[200200],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]%2){
          dp[x][0]+=dp[to[i]][1];
          dp[x][0]+=1ll*sum[to[i]][1]*val[i];
          sum[x][0]+=sum[to[i]][1];
          dp[x][1]+=dp[to[i]][0]+1ll*val[i];
          dp[x][1]+=1ll*sum[to[i]][0]*val[i];
          sum[x][1]+=sum[to[i]][0]+1;
      }else {
          dp[x][0]+=dp[to[i]][0]+1ll*val[i];
          dp[x][0]+=1ll*sum[to[i]][0]*val[i];
          sum[x][0]+=sum[to[i]][0]+1;
          dp[x][1]+=dp[to[i]][1];
          dp[x][1]+=1ll*sum[to[i]][1]*val[i];
          sum[x][1]+=sum[to[i]][1];
      }
    }
    return;
}
inline void dfs2(int x,int fa){
    int i,j,k;
    long long t0=0,t1=0;
    int s0=0,s1=0;
    for(i=head[x];i;i=nxt[i]){
      if(to[i]==fa)continue;
      if(val[i]%2){
          t0+=dp[to[i]][1];
          t0+=1ll*sum[to[i]][1]*val[i];
          s0+=sum[to[i]][1];
          t1+=dp[to[i]][0]+1ll*val[i];
          t1+=1ll*sum[to[i]][0]*val[i];
          s1+=sum[to[i]][0]+1;
      }else {
          t0+=dp[to[i]][0]+1ll*val[i];
          t0+=1ll*sum[to[i]][0]*val[i];
          s0+=sum[to[i]][0]+1;
          t1+=dp[to[i]][1];
          t1+=1ll*sum[to[i]][1]*val[i];
          s1+=sum[to[i]][1];
      }
    }
    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]%2){
          T0-=dp[to[i]][1];
          T0-=1ll*sum[to[i]][1]*val[i];
          S0-=sum[to[i]][1];
          T1-=dp[to[i]][0]+1ll*val[i];
          T1-=1ll*sum[to[i]][0]*val[i];
          S1-=sum[to[i]][0]+1;
          dp2[to[i]][0]+=dp2[x][1]+T1;
          dp2[to[i]][0]+=1ll*(sum2[x][1]+S1)*val[i];
          sum2[to[i]][0]+=sum2[x][1]+S1;
          dp2[to[i]][1]+=dp2[x][0]+1ll*val[i]+T0;
          dp2[to[i]][1]+=1ll*(sum2[x][0]+S0)*val[i];
          sum2[to[i]][1]+=sum2[x][0]+1+S0;
          dfs2(to[i],x);
      }else {
          T0-=dp[to[i]][0]+1ll*val[i];
          T0-=1ll*sum[to[i]][0]*val[i];
          S0-=sum[to[i]][0]+1;
          T1-=dp[to[i]][1];
          T1-=1ll*sum[to[i]][1]*val[i];
          S1-=sum[to[i]][1];
          dp2[to[i]][0]+=dp2[x][0]+1ll*val[i]+T0;
          dp2[to[i]][0]+=1ll*(sum2[x][0]+S0)*val[i];
          sum2[to[i]][0]+=sum2[x][0]+1+S0;
          dp2[to[i]][1]+=dp2[x][1]+T1;
          dp2[to[i]][1]+=1ll*(sum2[x][1]+S1)*val[i];
          sum2[to[i]][1]+=sum2[x][1]+S1;
          dfs2(to[i],x);
      }
    }
    return;
}
int main(){
    int i,j,k,x,y,z;
    scanf("%d%d",&n,&m);
    for(i=1;i<n;i++){
      scanf("%d%d%d",&x,&y,&z);
      add(x,y,z);
    }
    dfs1(1,0);
    dfs2(1,0);
    for(i=1;i<=m;i++){
      scanf("%d",&x);
      printf("%lld %lld
",dp[x][1]+dp2[x][1],dp[x][0]+dp2[x][0]);
    }
    return 0;
}