HDU6210 transaction transaction transaction

transaction transaction transaction

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 279    Accepted Submission(s): 123


Problem Description
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  ai  yuan in  i t city. Kelukin will take taxi, whose price is  1 yuan per km and this fare cannot be ignored.
There are  n1 roads connecting  n cities. Kelukin can choose any city to start his travel. He want to know the maximum money he can get.
 

Input
The first line contains an integer  T ( 1T10) , the number of test cases. 
For each test case:
first line contains an integer  n ( 2n100000) means the number of cities;
second line contains  n numbers, the  i th number means the prices in  i th city;  (1Price10000) 
then follows  n1 lines, each contains three numbers  x y and  z which means there exists a road between  x and  y, the distance is  z km  (1z1000)
 

Output
For 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
 

Source
2017 ACM/ICPC Asia Regional Shenyang Online

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题目的意思是给出一棵树,给出每个地点的买卖货物价格,和两个点之间的路费,求选出一个起点和一个终点,要求在起点买终点卖所得利润最大,求最大利润
思路:在原图基础上,建立一个源点和所有点连边,边权为每个点的价值,然后计算出源点到每个点的最短路,表示到这个点卖最少花费多少,然后遍历每个点算出获利求最大值
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

#define LL long long
const LL INF = 0x3f3f3f3f3f3f3f3f;

LL dis[100009],l[400009],w,a[100009];
int s[100009],nt[400009],e[400009],vis[100009];
int n,u,v;

struct node
{
    int id;
    LL dis;
    bool operator<(const node &a)const
    {
        return dis>a.dis;
    }
} pre,nt1;

void Dijkstra()
{
    memset(dis,INF,sizeof dis);
    memset(vis,0,sizeof vis);
    pre.id=pre.dis=0;
    dis[0]=0;
    priority_queueq;
    q.push(pre);
    while(!q.empty())
    {
        pre=q.top();
        q.pop();
        vis[pre.id]=1;
        for(int i=s[pre.id]; ~i; i=nt[i])
        {
            if(vis[e[i]]) continue;
            if(dis[e[i]]>pre.dis+l[i])
            {
                dis[e[i]]=pre.dis+l[i];
                nt1.id=e[i],nt1.dis=dis[e[i]];
                q.push(nt1);
            }
        }
    }
    LL ans=0;
    for(int i=1; i<=n; i++)
    {
        LL temp=max(a[i]-dis[i],1LL*0);
        ans=max(ans,temp);
    }
    printf("%lld\n",ans);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int cnt=0;
        memset(s,-1,sizeof s);
        for(int i=1; i<=n; i++)
        {
            scanf("%lld",&a[i]);
            nt[cnt]=s[i],s[i]=cnt,e[cnt]=0,l[cnt++]=a[i];
            nt[cnt]=s[0],s[0]=cnt,e[cnt]=i,l[cnt++]=a[i];
        }
        for(int i=1; i



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