hdu 6201 transaction transaction transaction (spfa求最长路)

transaction transaction transaction

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 1518    Accepted Submission(s): 733

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 aiyuan 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

 

题意:

给你n个点书在当地的价格，以及两个点之间距离的花费，问你求选两个点使的利润最大

解析：

新建两个点，一个源点，一个汇点，再用spfa求最长路

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<queue>using namespace std;const int MAXN = 500010 ;const int INF = 2000;typedef struct node{int u;int v;int value;int next;}node;node edge[MAXN];int head[MAXN],cnt,n;int d[MAXN],visit[MAXN];void addedge(int u,int v,int value){edge[cnt].u=u;edge[cnt].v=v;edge[cnt].value=value;edge[cnt].next=head[u];head[u]=cnt++;}void spfa(int s,int e){queue<int> mq;mq.push(s);visit[s]=1;d[s]=0;while(!mq.empty()){int tmp=mq.front();mq.pop();for(int i=head[tmp];i!=-1;i=edge[i].next){int v=edge[i].v;int value=edge[i].value;if(d[v]<d[tmp]+value){d[v]=d[tmp]+value;if(!visit[v]){mq.push(v);visit[v]=1;}}}visit[tmp]=0;}}int main(){int t;scanf("%d",&t);while(t--){memset(head,-1,sizeof(head));cnt=0;scanf("%d",&n);int a,b,c;for(int i=1;i<=n;i++){scanf("%d",&b);addedge(0,i,-b);addedge(i,n+1,b);}for(int i=1;i<=n-1;i++){scanf("%d%d%d",&a,&b,&c);addedge(a,b,-c);addedge(b,a,-c);}memset(visit,0,sizeof(visit));for(int i=0;i<=n+1;i++)d[i]=-INF;spfa(0,n+1);printf("%d\n",d[n+1]);}}