hdu 6201 transaction transaction transaction

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 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 2 301 3 23 4 10

 

Sample Output

8

 

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]);    }}