hdu 6201 transaction transaction transaction

最新推荐文章于 2018-10-23 09:39:33 发布

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本文介绍了一个基于图论的算法问题，旨在寻找商人Kelukin在不同城市间买卖书籍以获得最大利润的最优路径。通过构建特殊的源点和汇点，并利用SPFA算法，解决了带有成本的交易问题。

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transaction transaction transaction

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 1244 Accepted Submission(s): 595

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

Recommend

liuyiding

最长路。。

建立两个新的点。

一个叫做源点，一个是汇点。

然后在这两点间最长的路就是答案了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define inf 0x3f3f3f3f
#define N 100005
#define mem(a,x) memset(a,x,sizeof(a))
struct data
{
    int to;
    int cost;
};
int n;
int dis[N];
int vis[N];
int Point[N];
vector<data>v[N];
void init()
{
    for(int i=0; i<N; i++)
        v[i].clear();
}
int add(int a,int b,int c)
{
    data tmp;
    tmp.cost=c;
    tmp.to=b;
    v[a].push_back(tmp);
}
void spfa(int u)
{
    memset(vis,0,sizeof(vis));
    memset(dis,-inf,sizeof(dis));
    queue<int>q;
    vis[u]=1;
    q.push(u);
    dis[u]=0;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        vis[u]=0;
        int len=v[u].size();
        for(int i=0; i<len; i++)
        {
            int to=v[u][i].to;
            int cost=v[u][i].cost;
            int tmp=dis[u]+cost;
            if(tmp>dis[to])
            {
                dis[to]=tmp;
                if(!vis[to])
                {
                    vis[to]=1;
                    q.push(to);
                }
               // printf("--%d\n",u);
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&Point[i]);
        }
        Point[0]=Point[n+1]=0;
        for(int i=1; i<=n; i++)
        {
            add(0,i,-2);///为新建源点与各点的边赋值
            add(i,n+1,2);///为新建汇点与各点的边赋值  距离随便填了，，只要对称就好
        }
        int x,y,c;
        for(int i=1; i<n; i++)
        {
            scanf("%d%d%d",&x,&y,&c);
            add(x,y,Point[y]-Point[x]-c);
            add(y,x,Point[x]-Point[y]-c);
        }
        spfa(0);
        //printf("---------\n");
        /*for(int i=0;i<=n+1;i++)
        {
            printf("%d ",dis[i]);
        }
        printf("\n");*/
        printf("%d\n",dis[n+1]);
    }
}