TOJ 2967: Job Hunt -- Bellman_Ford

最新推荐文章于 2019-12-30 21:17:58 发布

原创最新推荐文章于 2019-12-30 21:17:58 发布 · 880 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#图 #单源最短路 #负权回路 #Bellman_Ford #dp

ACM 专栏收录该内容

56 篇文章

订阅专栏

Bessie在有限时间内通过遍历多个城市寻找工作机会，每次可在每个城市赚取固定金额，利用路径和 Farmer John 提供的飞行服务，目标是在遵循规则的情况下赚取最大金额。

2967: Job Hunt

Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit: 9 Accepted:7

Description

Bessie is running out of money and is searching for jobs. Farmer John knows this and wants the cows to travel around so he has imposed a rule that his cows can only make D (1 ≤ D ≤ 1,000) dollars in a city before they must work in another city. Bessie can, however, return to a city after working elsewhere for a while and again earn the D dollars maximum in that city. There is no limit on the number of times Bessie can do this.

Bessie's world comprises P (1 ≤ P ≤ 150) one-way paths connecting C (2 ≤ C ≤ 220) cities conveniently numbered 1..C. Bessie is currently in city S (1 ≤ S ≤ C). Path i runs one-way from city A_i to city B_i (1 ≤ A_i ≤ C; 1 ≤ B_i ≤ C) and costs nothing to traverse.

To help Bessie, Farmer John will give her access to his private jet service. This service features F (1 ≤ F ≤ 350) routes, each of which is a one way flight from one city J_i to a another K_i (1 ≤ J_i ≤ C; 1 ≤ K_i ≤ C) and which costs T_i (1 ≤ T_i ≤ 50,000) dollars. Bessie can pay for the tickets from future earnings if she doesn't have the cash on hand.

Bessie can opt to retire whenever and wherever she wants. Given an unlimited amount of time, what is the most money that Bessie can make presuming she can make the full D dollars in each city she can travel to? Print -1 if there is no limit to this amount.

Input

* Line 1: Five space-separated integers: D, P, C, F, and S

* Lines 2..P+1: Line i+1 contains two space-separated integers that name a one-way path from one city to another: A_i and B_i

* Lines P+2..P+F+1: Line P+i+1 contains three space-separated integers that name a one-way jet flight from one city to another and the price for that flight: J_i, K_i, and T_i

Output

* Line 1: A single integer representing the most money she can make while following the law.

Sample Input

100 3 5 2 1
1 5
2 3
1 4
5 2 150
2 5 120

Sample Output

Source

USACO Nov 2009

分析：带负权回路的单源最短路。

#include 
    
     
#include 
     
      
#include 
      
       
#include 
       
        
#include 
        
         
#include 
          #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include 
              
                #include 
               
                 #include 
                
                  #include 
                 
                   #include 
                  
                    #include 
                   
                     #define mp make_pair using namespace std; typedef unsigned int ui; typedef long long ll; typedef unsigned long long ull; typedef pair
                    
                      pii; typedef vector
                     
                       vi; typedef vi::iterator vi_it; typedef map
                      
                        mii; typedef priority_queue
                       
                         pqi; typedef priority_queue
                        
                         , greater
                         
                           > rpqi; const int MAX_C = 220 + 2; const int MAX_P = 500 + 2; const int INF = 500000000; int dp[MAX_C]; int cnt = 0; struct edge { int u; int v; int w; } e[MAX_P]; class Bellman_Ford { public: Bellman_Ford(int _n, int _s); bool bellman_ford(); private: int n, s; void initialize_single_source(); inline void relax(int u, int v, int duv); }; Bellman_Ford::Bellman_Ford(int _n, int _s) : n(_n), s(_s) { } void Bellman_Ford::initialize_single_source() { for (int i = 1; i <= n; ++i) { dp[i] = -INF; } dp[s] = 0; } inline void Bellman_Ford::relax(int u, int v, int duv) { if (dp[v] < dp[u] + duv) { dp[v] = dp[u] + duv; } } bool Bellman_Ford::bellman_ford() { initialize_single_source(); for (int k = 1; k < n; ++k) { for (int i = 0; i < cnt; ++i) { relax(e[i].u, e[i].v, e[i].w); } } for (int i = 1; i < cnt; ++i) { if (dp[e[i].v] < dp[e[i].u] + e[i].w) { return false; } } return true; } int main(int argc, char *argv[]) { // freopen("D:\\in.txt", "r", stdin); int d, p, c, f, s; cin >> d >> p >> c >> f >> s; while (p--) { int a, b; cin >> a >> b; e[cnt].u = a; e[cnt].v = b; e[cnt++].w = d; } while (f--) { int j, k, t; cin >> j >> k >> t; e[cnt].u = j; e[cnt].v = k; e[cnt++].w = d - t; } Bellman_Ford bf(c, s); if (bf.bellman_ford()) { int maxv = 0; for (int k = 1; k <= c; ++k) { maxv = max(maxv, dp[k]); } cout << maxv + d << endl; } else { cout << -1 << endl; } return 0; }