HDU6201(spfa)

最新推荐文章于 2020-02-12 17:52:53 发布

creatorx

最新推荐文章于 2020-02-12 17:52:53 发布

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分类专栏：最短路

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transaction transaction transaction

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 842 Accepted Submission(s): 411

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

解题思路：我们增加一个源点，让所有点与这个源点相连，边权为a[i], 增加一个汇点，让所有点与这个汇点相连，边权为-a[i],然后我们求一条从源点到汇点的最长路径就行，因为涉及到负边，所以我们用spfa就行。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 100;
int n;
int d[maxn];
bool visit[maxn];
int inf = 0x3f3f3f3f;
struct node{
    int v;
    int w;
    node(int _v = 0, int _w = 0){
        v = _v;
        w = _w;
    }
};
queue<node> q;
vector<pair<int ,int> > g[maxn];
int spfa()
{
   memset(d, -inf, sizeof(d));
   memset(visit, false, sizeof(visit));
   while(!q.empty()) q.pop();
   d[0] = 0;
   q.push(node(0, 0));
   while(!q.empty())
   {
       node nx = q.front();
       q.pop();
       int v = nx.v;
       visit[v] = false;
       for(int i = 0; i < g[v].size(); i++)
       {
           int u = g[v][i].first;
           int ww = g[v][i].second;
           if(d[v] + ww > d[u] && u != 0)
           {
               d[u] = d[v] + ww;
               if(visit[u]) continue;
               visit[u] = true;
               q.push(node(u, d[u]));
           }
       }
   }
   if(d[n + 1] > 0) return d[n + 1];
   else return 0;
}
void init()
{
    for(int i = 0; i <= n + 1; i++)
    {
        g[i].clear();
    }
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        init();
        int u, v, w;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &w);
            g[0].push_back(make_pair(i, w));
            g[i].push_back(make_pair(0, w));
            g[n + 1].push_back(make_pair(i, -w));
            g[i].push_back(make_pair(n + 1, -w));
        }
        for(int i = 1; i < n; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            g[u].push_back(make_pair(v, -w));
            g[v].push_back(make_pair(u, -w));
        }
        printf("%d\n", spfa());
    }
    return 0;
}