SGU 143 Long Live the Queen （樹形DP）

143. Long Live the Queen

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

The Queen of Byteland is very loved by her people. In order to show her their love, the Bytelanders have decided to conquer a new country which will be named according to the queen's name. This new country contains N towns. The towns are connected by bidirectional roads and there is exactly ONE path between any two towns, walking on the country's roads. For each town, the profit it brings to the owner is known. Although the Bytelanders love their queen very much, they don't want to conquer all the N towns for her. They will be satisfied with a non-empty subset of these towns, with the following 2 properties: there exists a path from every town in the subset to every other town in the subset walking only through towns in the subset and the profit of the subset is maximum. The profit of a subset of the N towns is equal to the sum of the profits of the towns which belong to the subset. Your task is to find the maximum profit the Bytelanders may get.

Input

The first line of input will contain the number of towns N (1<=N<=16 000). The second line will contain N integers: the profits for each town, from 1 to N. Each profit is an integer number between-1000 and 1000. The next N-1 lines describe the roads: each line contains 2 integer numbers a and b, separated by blanks, denoting two different towns between which there exists a road.

Output

The output should contain one integer number: the maximum profit the Bytelanders may get.

Sample Input

5
-1 1 3 1 -1
4 1
1 3
1 2
4 5

Sample Output

4

題意 ：給定一棵樹，每個節點上權值爲整數。求它的一個連通子圖，使其所有節點權值之和最大。

思路 ：dp[ i ] 表示以編號 i 的節點爲根的字數中，在選 i 這個點的前提下，可以選擇的最大子樹的權值和。

即   dp [ i ] += max(0,dp[ v ] )  v 爲節點 i的兒子節點。 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int inf=999999999;
const int maxn=16010;

int dp[maxn],a[maxn],n,root;
vector <int> G[maxn];

void input()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)  scanf("%d",&a[i]);
    int u,v;
    for(int i=1; i<n; i++)
    {
        scanf("%d %d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
}

void dfs(int u,int v)
{
    for(int i=0; i<G[v].size(); i++)
    {
        int vv=G[v][i];
        if(vv!=u)  dfs(v,vv);
    }
    for(int i=0; i<G[v].size(); i++)
    {
        int vv=G[v][i];
        if(vv!=u) dp[v]+=max(0,dp[vv]);
    }
}

void solve()
{
    for(int i=1; i<=n; i++) dp[i]=a[i];
    dfs(-1,1);
    int ans=-inf;
    for(int i=1; i<=n; i++)  ans=max(ans,dp[i]);
    printf("%d\n",ans);
}

int main()
{
    input();
    solve();
    return 0;
}