Nowcoder-2018ACM多校訓練營（第二場）H travel

時間限制：C/C++ 2秒，其他語言4秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld

題目描述 

White Cloud has a tree with n nodes.The root is a node with number 1. Each node has a value.
White Rabbit wants to travel in the tree 3 times. In Each travel it will go through a path in the tree.
White Rabbit can't pass a node more than one time during the 3 travels. It wants to know the maximum sum value of all nodes it passes through.

 

 

輸入描述:

The first line of input contains an integer n(3 <= n <= 400001)
In the next line there are n integers in range [0,1000000] denoting the value of each node.
For the next n-1 lines, each line contains two integers denoting the edge of this tree.

輸出描述:

Print one integer denoting the answer.

示例1

輸入

複製

13
10 10 10 10 10 1 10 10 10 1 10 10 10
1 2
2 3
3 4
4 5
2 6
6 7
7 8
7 9
6 10
10 11
11 12
11 13

輸出

複製

110

題意：在樹上找3條不相交鏈，要求鏈上權值和最大

題解：樹型dp，設dp[u][j][k]：以u爲頂點的子樹上有j條不相交鏈，k=0表示頂點不可被父節點連接，k=1表示頂點可被父節點連接。一堆轉移方程，寫的時候漏掉好幾個，WA到想吐血系列。。。

#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;

typedef long long LL;
typedef pair<int, int> PII;

const int MX = 1e6 + 5;

struct Edge {
    int v, nxt;
} E[MX * 2];
int head[MX], tot;

void init (int n) {
    for (int i = 1; i <= n; i++) head[i] = -1;
    tot = 0;
}

void edge_add (int u, int v) {
    E[tot] = (Edge) {v, head[u]}; head[u] = tot++;
}

int a[MX];
LL dp[MX][4][2], t[MX][4][2];

inline LL Max (LL d[2]) {
    return max (d[0], d[1]);
}

void dfs (int u, int pre) {
    memset (dp[u], 0, sizeof (dp[u]) );
    memset (t[u], 0, sizeof (t[u]) );
    dp[u][1][1] = a[u];
    for (int i = head[u]; ~i; i = E[i].nxt) {
        int v = E[i].v;
        if (v == pre) continue;
        dfs (v, u);
        dp[u][3][0] = max (dp[u][3][0], Max (dp[v][3]) );
        dp[u][3][0] = max (dp[u][3][0], Max (dp[v][2]) + Max (t[u][1]) );
        dp[u][3][0] = max (dp[u][3][0], Max (dp[v][1]) + Max (t[u][2]) );
        dp[u][3][0] = max (dp[u][3][0], a[u] + t[u][1][1] + dp[v][3][1]);
        dp[u][3][0] = max (dp[u][3][0], a[u] + t[u][2][1] + dp[v][2][1]);
        dp[u][3][0] = max (dp[u][3][0], a[u] + t[u][3][1] + dp[v][1][1]);
        dp[u][3][0] = max (dp[u][3][0], dp[u][2][0] + Max (dp[v][1]) );
        dp[u][3][0] = max (dp[u][3][0], dp[u][1][0] + Max (dp[v][2]) );
        dp[u][3][1] = max (dp[u][3][1], a[u] + dp[v][3][1]);
        dp[u][3][1] = max (dp[u][3][1], a[u] + max (dp[v][2][1] + Max (t[u][1]), Max (dp[v][2]) + t[u][1][1]) );
        dp[u][3][1] = max (dp[u][3][1], a[u] + max (dp[v][1][1] + Max (t[u][2]), Max (dp[v][1]) + t[u][2][1]) );
        dp[u][3][1] = max (dp[u][3][1], a[u] + Max (dp[v][2]) );
        dp[u][3][1] = max (dp[u][3][1], a[u] + Max (dp[v][1]) + Max (t[u][1]) );

        dp[u][2][0] = max (dp[u][2][0], Max (dp[v][2]) );
        dp[u][2][0] = max (dp[u][2][0], Max (dp[v][1]) + Max (t[u][1]) );
        dp[u][2][0] = max (dp[u][2][0], a[u] + t[u][1][1] + dp[v][2][1]);
        dp[u][2][0] = max (dp[u][2][0], a[u] + t[u][2][1] + dp[v][1][1]);
        dp[u][2][0] = max (dp[u][2][0], dp[u][1][0] + Max (dp[v][1]) );
        dp[u][2][1] = max (dp[u][2][1], a[u] + dp[v][2][1]);
        dp[u][2][1] = max (dp[u][2][1], a[u] + max (dp[v][1][1] + Max (t[u][1]), Max (dp[v][1]) + t[u][1][1]) );
        dp[u][2][1] = max (dp[u][2][1], a[u] + Max (dp[v][1]) );

        dp[u][1][0] = max (dp[u][1][0], Max (dp[v][1]) );
        dp[u][1][0] = max (dp[u][1][0], a[u] + dp[v][1][1] + t[u][1][1]);
        dp[u][1][1] = max (dp[u][1][1], a[u] + dp[v][1][1]);


        t[u][3][0] = max (t[u][3][0], Max (dp[v][3]) );
        t[u][3][0] = max (t[u][3][0], Max (dp[v][2]) + Max (t[u][1]) );
        t[u][3][0] = max (t[u][3][0], Max (dp[v][1]) + Max (t[u][2]) );
        t[u][3][1] = max (t[u][3][1], dp[v][3][1]);
        t[u][3][1] = max (t[u][3][1], max (dp[v][2][1] + Max (t[u][1]), Max (dp[v][2]) + t[u][1][1]) );
        t[u][3][1] = max (t[u][3][1], max (dp[v][1][1] + Max (t[u][2]), Max (dp[v][1]) + t[u][2][1]) );

        t[u][2][0] = max (t[u][2][0], Max (dp[v][2]) );
        t[u][2][0] = max (t[u][2][0], Max (dp[v][1]) + Max (t[u][1]) );
        t[u][2][1] = max (t[u][2][1], dp[v][2][1]);
        t[u][2][1] = max (t[u][2][1], max (dp[v][1][1] + Max (t[u][1]), Max (dp[v][1]) + t[u][1][1]) );

        t[u][1][0] = max (t[u][1][0], Max (dp[v][1]) );
        t[u][1][1] = max (t[u][1][1], dp[v][1][1]);
    }
}

int main() {
    int n;
    //freopen ("in.txt", "r", stdin);
    while (~scanf ("%d", &n) ) {
        init (n);
        for (int i = 1; i <= n; i++) scanf ("%d", &a[i]);
        for (int i = 1; i < n; i++) {
            int u, v;
            scanf ("%d%d", &u, &v);
            edge_add (u, v);
            edge_add (v, u);
        }
        dfs (1, -1);

        printf ("%lld\n", Max (dp[1][3]) );
    }
    return 0;
}