51nod 1424 零树(树形dp)

题目：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1424
思路：与上题相似，dp[u][1/0]代表将结点u所在联通块情况所需要的+1/-1的操作数
dp[u][1] = max(dp[son][1])
dp[u][0] = max(dp[son][0])
然后再处理下情况所需要的差值
代码：

#include<bits/stdc++.h>using namespace std;const int N = 1e5+5;int v[N];long long dp[N][2];vector<int> G[N];int n;int dfs(int u,int fa){    for(int i = 0;i < G[u].size();i++)    {        int son = G[u][i];        if(son == fa)            continue;        dfs(son,u);        dp[u][1] = max(dp[u][1],dp[son][1]);        dp[u][0] = max(dp[u][0],dp[son][0]);    }    int diff = v[u] + dp[u][1] - dp[u][0];    dp[u][diff < 0] += abs(diff);}int main(){    scanf("%d",&n);    int a,b;    for(int i = 0;i < n-1;i++)    {        scanf("%d%d",&a,&b);        G[a].push_back(b);        G[b].push_back(a);    }    for(int i = 1;i <= n;i++)        scanf("%d",&v[i]);    dfs(1,-1);    printf("%lld\n",dp[1][0]+dp[1][1]);    return 0;}