HDU - 6201 transaction transaction transaction（樹形dp取兩點）

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





一我的在任意點買書（消費點權），通過邊（花費邊權），在任意點賣書（收穫點權），求最大收益。
樹形結構，由於父子關係未知，因此雙向建邊。
dp[i][0]表示i子樹中的最少買書消費，dp[i][1]表示i子樹中的最大賣書收益，
dp[i][0]+dp[i][1]表示以i爲最近公共父節點的最大收益，最優解並不是根節點，所以ans須要不斷更新。

#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]);     //不斷更新
}
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;
}