loj10160. 「一本通 5.2 练习 3」周年纪念晚会

思路：
　　挺简单的树形dp

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<vector>
using namespace std;
const int maxn = 200010;
void qread(int &x) {
    x = 0;
    register int ch = getchar(), flag = 0;
    while(ch < '0' || ch > '9')    {if(ch == '-')    flag = 1;    ch = getchar();}
    while(ch >= '0' && ch <= '9')    x = 10 * x + ch - 48, ch = getchar();
    if(flag)    x = -x;
}
int n, rt = 0, cnt;

int val[maxn];
int deep[maxn];
int f[maxn];

int head[maxn];
int go[maxn << 1];
int nxt[maxn << 1];

int dp[maxn][2];

inline void init(){
    qread(n);
    for(int i=1; i<=n; ++i)
        qread(val[i]);
    int x, y;
    qread(x), qread(y);
    while(x || y){
        go[++cnt] = x;
        nxt[cnt] = head[y];
        head[y] = cnt;
        f[x] = y;
        qread(x), qread(y);
    }
    for(int i=1; i<=n; ++i)
        if(!f[i])
            rt = i;    
}

void DP(int x){
    dp[x][1] = val[x];
    for(int i = head[x]; i; i = nxt[i]){
        DP(go[i]);
        dp[x][1] += dp[go[i]][0];
        dp[x][0] += max(dp[go[i]][0], dp[go[i]][1]);
    }
}
int main(void){
    init();
    DP(rt);
    printf("%d\n", max(dp[rt][0], dp[rt][1]));
}

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