HDU 6201 transaction transaction transaction （2017沈阳网络赛 - spfa最长路）

题意：

给你一棵树， 树上有点权， 要求选择起点S和终点T, 要求T-S-sum 最大， sum为S到T的边权。

思路：

根据题意就可以建图

建立源点和汇点。

源点连所有的树上点， 边权为 a[i], 所有树上点在连接 汇点， 边权为-a[i]. 然后在根据树建图。 

spfa跑个最长路即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
int T;

const int maxn = 100000 + 10;

int d[maxn];
int a[maxn];
struct Edge{
    int u, v, d;
    Edge(int u = 0,int v = 0,int d = 0):u(u), v(v), d(d){}
};
vector<Edge>edges;
int n;
vector<int>g[maxn];
void add(int u,int v,int w){
    edges.push_back(Edge(u, v, w));
    g[u].push_back((int)edges.size() - 1);
}
bool vis[maxn];
int cnt[maxn];
queue<int>q;
void spfa(){
    memset(vis,0,sizeof vis);
    memset(cnt,0,sizeof cnt);
    while(!q.empty())q.pop();
    vis[0] = 1; cnt[0] = 1;
    memset(d,0,sizeof d);
//    d[0] = 0;
    q.push(0);
    while(!q.empty()){
        int u = q.front(); q.pop();
        vis[u] = 0;
        for (int i = 0; i < g[u].size(); ++i){
            int id = g[u][i];
            int v = edges[id].v;
            int w = edges[id].d;
            if (d[v] < d[u] + w){
                d[v] = d[u] + w;
                if (!vis[v]){
                    vis[v] = 1;
                    cnt[v]++;
                    q.push(v);
                    if (cnt[v] > n) return;
                }
            }
        }
    }
    printf("%d\n", d[n+1]);
}
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        edges.clear();
        for (int i = 1; i <= n; ++i){
            scanf("%d", &a[i]);
            g[i].clear();
        }

        for (int i = 1; i <= n; ++i){
            add(0, i, a[i]);
            add(i, n+1, -a[i]);
        }
        for (int i = 1; i < n; ++i){
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            add(u, v, -w);
            add(v, u, -w);
        }
        spfa();
    }

    return 0;
}

transaction transaction transaction

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 485    Accepted Submission(s): 225

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  i t city. Kelukin will take taxi, whose price is  1 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

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  i th number means the prices in  i 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  z km  (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

8

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

Recommend

liuyiding   |   We have carefully selected several similar problems for you:   6205  6204  6203  6202  6201