牛客网暑期ACM多校训练营（第二场）: H. travel（树形线头DP）

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

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;
}