143. Long Live the Queen 树形dp 难度:0

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 and1000. 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

题意:求一棵收益最大的树/子树,不能为空
思路:分别对每个节点维护以该节点为根所能得到的最大收益,更新答案即可
转移方程dp[i]=sum(dp[son[i]]>0?dp[son[i]]:0)

#include <cstdio>
#include <cstring>
#include<algorithm>
using namespace std;
const int maxn=16001;
const int maxm=32001;
int first[maxn],next[maxm],to[maxm],profit[maxn],len,n;
int sum[maxn];
void addedge(int f,int t){
    next[len]=first[f];
    first[f]=len;
    to[len]=t;
    swap(f,t);len++;
    next[len]=first[f];
    first[f]=len;
    to[len]=t;
    len++;
}
int dfs(int s,int f){
    sum[s]=profit[s];
    for(int p=first[s];p!=-1;p=next[p]){
        int t=to[p];
        if(t==f)continue;
        int son=dfs(t,s);
        if(son>0)sum[s]+=son;
    }
    return sum[s];
}
int main(){
    scanf("%d",&n);
    int tf,tt;
    memset(first,-1,sizeof(first));
    for(int i=1;i<=n;i++)scanf("%d",profit+i);
    for(int i=1;i<n;i++){scanf("%d%d",&tf,&tt);addedge(tf,tt);}
    dfs(1,-1);
    int maxn=-0x7ffffff;
    for(int i=1;i<=n;i++){maxn=max(maxn,sum[i]);}
    printf("%d\n",maxn);
}