HDU - 6201 transaction transaction transaction（树形dp取两点）

transaction transaction transaction

 

Kelukin is a businessman. Every day, he travels around cities to do some business. On August 17th, in memory of a great man, citizens will read a book named "the Man Who Changed China". Of course, Kelukin wouldn't miss this chance to make money, but he doesn't have this book. So he has to choose two city to buy and sell. 
As we know, the price of this book was different in each city. It is  aiai yuanyuan in iittcity. Kelukin will take taxi, whose price is 11yuanyuan per km and this fare cannot be ignored. 
There are n−1n−1 roads connecting nn cities. Kelukin can choose any city to start his travel. He want to know the maximum money he can get. 

InputThe first line contains an integer TT (1≤T≤101≤T≤10) , the number of test cases. 
For each test case: 
first line contains an integer nn (2≤n≤1000002≤n≤100000) means the number of cities; 
second line contains nn numbers, the iithth number means the prices in iithth city; (1≤Price≤10000)(1≤Price≤10000) 
then follows n−1n−1 lines, each contains three numbers xx, yy and zz which means there exists a road between xx and yy, the distance is zzkmkm (1≤z≤1000)(1≤z≤1000). 
OutputFor each test case, output a single number in a line: the maximum money he can get. 
Sample Input

1  
4  
10 40 15 30  
1 2 30
1 3 2
3 4 10

Sample Output

8





一我的在任意点买书（消费点权），通过边（花费边权），在任意点卖书（收获点权），求最大收益。
树形结构，由于父子关系未知，因此双向建边。
dp[i][0]表示i子树中的最少买书消费，dp[i][1]表示i子树中的最大卖书收益，
dp[i][0]+dp[i][1]表示以i为最近公共父节点的最大收益，最优解并不是根节点，所以ans须要不断更新。

#include<bits/stdc++.h>
#define MAX 100005
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

int a[MAX];
int dp[MAX][2];
int ans;
struct Node{
    int v,w;
}node;
vector<Node> v[MAX];

void dfs(int x,int pre){
    dp[x][0]=-a[x];dp[x][1]=a[x];
    for(int i=0;i<v[x].size();i++){
        int to=v[x][i].v;
        if(to==pre) continue;
        int w=v[x][i].w;
        dfs(to,x);
        dp[x][0]=max(dp[x][0],dp[to][0]-w);
        dp[x][1]=max(dp[x][1],dp[to][1]-w);
    }
    ans=max(ans,dp[x][0]+dp[x][1]);     //不断更新
}
int main()
{
    int t,n,i;
    int x,y,w;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            v[i].clear();
        }
        for(i=1;i<n;i++){
            scanf("%d%d%d",&x,&y,&w);
            node.v=y;node.w=w;
            v[x].push_back(node);
            node.v=x;
            v[y].push_back(node);
        }
        ans=-INF;
        dfs(1,-1);
        printf("%d\n",ans);
    }
    return 0;
}