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 n1 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 (1T10) , the number of test cases. 
For each test case:
first line contains an integer n (2n100000) means the number of cities;
second line contains n numbers, the ith number means the prices in ith city; (1Price10000) 
then follows n1 lines, each contains three numbers xy and z which means there exists a road between x and y, the distance is zkm (1z1000)
 

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
 

Recommend
liuyiding


最長路 。。
建立兩個新的點。
一個叫做源點,一個是匯點。
然後在這兩點間最長的路就是答案了。
#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);///爲新建源點與各點的邊賦值
            add(i,n+1,2);///爲新建匯點與各點的邊賦值  距離隨便填了,,只要對稱就好
        }
        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]);
    }
}


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