HDU6210 transaction transaction transaction_hdu

本文介绍了一个基于图论的问题，即如何在一个由多个城市组成的网络中找到最佳的买卖路径以获得最大利润。通过构建一个加权图，并利用Dijkstra算法计算从虚拟源点到各城市的最短路径，进而确定最佳买卖策略。

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

yuan in

t city. Kelukin will take taxi, whose price is

yuan per km and this fare cannot be ignored.
There are

n−1 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 (

1≤T≤10 ) , the number of test cases.
For each test case:
first line contains an integer

n (

2≤n≤100000 ) means the number of cities;
second line contains

n numbers, the

th number means the prices in

th city;

(1≤Price≤10000)
then follows

n−1 lines, each contains three numbers

x ,

y and

z which means there exists a road between

x and

y , the distance is

(1≤z≤1000) .

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

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 <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

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_queue<node>q;
    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<n; i++)
        {
            scanf("%d%d%lld",&u,&v,&w);
            nt[cnt]=s[u],s[u]=cnt,e[cnt]=v,l[cnt++]=w;
            nt[cnt]=s[v],s[v]=cnt,e[cnt]=u,l[cnt++]=w;
        }
        Dijkstra();
    }
    return 0;
}