bzoj2060

树形dp

dp[x][0]表示x点父亲没选，dp[x][1]表示x点父亲选了，然后dp[x][0]=max(sigma(dp[c[x]][0]),sigma(dp[c[x]][1])) dp[x][1]=sigma(dp[c[x][0]])

答案就是dp[1][0] 根没有父亲

#include<bits/stdc++.h>
using namespace std;
const int N = ;
int n,cnt = ;
int head[N], dp[N][];
struct edge {
    int nxt, to;
} e[N << ];
inline void link(int u, int v)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
}
void dfs(int u, int last)
{
    int sum_0 = , sum_1 = ;
    for(int i = head[u]; i; i = e[i].nxt)
    {
        if(e[i].to == last) continue;
        dfs(e[i].to, u);
        sum_0 += dp[e[i].to][];
        sum_1 += dp[e[i].to][];
    }
    dp[u][] = max(sum_0, sum_1 + );
    dp[u][] = sum_0;
}
int main()
{
    scanf("%d", &n);
    for(int i = ; i < n; ++i)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        link(u, v);
        link(v, u);
    }
    dfs(, );
    printf("%d\n", dp[][]);
    return ;
}