HDU 6201 transaction transaction transaction（网络流+最短路）

transaction transaction transaction

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 it city. Kelukin will take taxi, whose price is 1yuan 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 ith number means the prices in ith 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 zkm (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

8

Source

2017 ACM/ICPC Asia Regional Shenyang Online

题意：给出n个结点的一棵树，选择两个结点买书和卖书，每条路径有花费，求赚的最大钱数。

题解：网络流思想，创建源点，与n个结点相连，cost为书的价格，表示买书的花费；创建汇点，与n个结点相连，cost为负的书的价格，表示卖书赚的钱。

因为最大流为1，所以可用最短路（sfpa）求出源点到汇点的最短距离，也就是所赚钱数的负值。

#include<iostream>#include<stdio.h>#include<string.h>#include<vector>#include<algorithm>#include<queue>#define ll long long#define inf 1e9+5#define maxn 100005using namespace std;struct edge{    int from,to,w,next;}e[500005];int head[maxn];int vis[maxn];int dist[maxn];int n,m,t;void add(int i,int j,int w){    e[t].from=i;    e[t].to=j;    e[t].w=w;    e[t].next=head[i];    head[i]=t++;}void spfa(int s){    queue <int> q;    for(int i=0;i<n;i++)    dist[i]=inf;    memset(vis,false,sizeof(vis));    q.push(s);    dist[s]=0;    while(!q.empty())    {        int u=q.front();        q.pop();        vis[u]=false;        for(int i=head[u];i!=-1;i=e[i].next)        {            int v=e[i].to;            if(dist[v]>dist[u]+e[i].w)            {                dist[v]=dist[u]+e[i].w;                if(!vis[v])                {                    vis[v]=true;                    q.push(v);                }            }        }    }}int a[100005];int main(){    int tt;    scanf("%d",&tt);    while(tt--)    {        int nn;        scanf("%d",&nn);        n=nn+2;        t=0;        memset(head,-1,sizeof(head));        for(int i=1;i<=nn;i++)          {              scanf("%d",&a[i]);              add(0,i,a[i]);              add(i,nn+1,-a[i]);          }        int u,v,w;        for(int i=1;i<nn;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,w);            add(v,u,w);        }        spfa(0);        cout<<-dist[nn+1]<<endl;    }    return 0;}