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链接：https://ac.nowcoder.com/acm/contest/140/H
来源：牛客网
题目描述
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.
输入描述:
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.
输出描述:
Print one integer denoting the answer.
输入
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
输出
110
题意：
给你一棵n个节点的树，每个点都有一个权值，选出三条不相交路径使得权值和最大
思路：
dp[i][j][0]表示以i为根的子树选了j条路径的最大权值和
dp[i][j][1]表示以i为根的子树选了j条路径，其中一条路径的一端为i的最大权值和
那么dp时来一波类似于背包的转移就好了
//https://ac.nowcoder.com/acm/contest/140/H
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
#define LL long long
#define mod 1000000007
vector<int> G[500005];
int val[500005];
LL dp[500005][5][2];
void Sech(int u, int fa)
{
int i, j, p, q, v, t;
LL now[5][4], c[5][4], sum[5][4];
memset(now, 0, sizeof(now));
for(t=0;t<G[u].size();t++)
{
v = G[u][t];
if(v==fa)
continue;
Sech(v, u);
memset(c, 0, sizeof(c));
for(i=0;i<=3;i++)
c[i][0] = dp[v][i][0], c[i][1] = dp[v][i][1];
memset(sum, 0, sizeof(sum));
for(i=0;i<=3;i++)
{
for(j=0;i+j<=3;j++)
{
for(p=0;p<=2;p++)
{
for(q=0;p+q<=2;q++)
sum[i+j][p+q] = max(sum[i+j][p+q], now[i][p]+c[j][q]);
}
}
}
memcpy(now, sum, sizeof(now));
}
for(i=0;i<=3;i++)
dp[u][i][0] = now[i][0];
for(i=0;i<=2;i++)
{
for(j=0;j<=2;j++)
dp[u][i+1][0] = max(dp[u][i+1][0], now[i][j]+val[u]);
}
for(i=0;i<=3;i++)
{
for(j=0;j<=1;j++)
dp[u][i][1] = max(dp[u][i][1], now[i][j]+val[u]);
}
}
int main(void)
{
int n, i, x, y;
scanf("%d", &n);
for(i=1;i<=n;i++)
scanf("%d", &val[i]);
for(i=1;i<=n-1;i++)
{
scanf("%d%d", &x, &y);
G[x].push_back(y);
G[y].push_back(x);
}
Sech(1, 0);
printf("%lld\n", dp[1][3][0]);
return 0;
}