poj3159 Candies

Posted 2020-08-29 王宜鸣

题意：

有N个孩子（N<=3000)分糖果。
有M个关系(M<=150,000)。每个关系形如：
A B C
表示第B个学生比第A个学生多分到的糖果数目，不能超过C
求第N个学生最多比第1个学生能多分几个糖果

思路：

dijkstra+heap

实现：

 1 #include <iostream>
 2 #include <vector>
 3 #include <queue>
 4 #include <functional>
 5 #include <cstdio>
 6 using namespace std;
 7 
 8 const int INF = 0x3f3f3f3f;
 9 
10 struct edge
11 {
12     int to;
13     int cost;
14 };
15 vector<edge> G[30005];
16 int d[30005];
17 typedef pair<int, int>P;
18 
19 int Scan()
20 {
21     int res = 0, ch, flag = 0;
22 
23     if ((ch = getchar()) == ‘-‘)             
24         flag = 1;
25 
26     else if (ch >= ‘0‘ && ch <= ‘9‘)          
27         res = ch - ‘0‘;
28     while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘)
29         res = res * 10 + ch - ‘0‘;
30 
31     return flag ? -res : res;
32 }
33 
34 int main()
35 {
36     int n, m, a, b, l;
37     cin >> n >> m;
38     for (int i = 0; i < m; i++)
39     {
40         a = Scan();
41         b = Scan();
42         l = Scan();
43         edge e;
44         e.to = b;
45         e.cost = l;
46         G[a].push_back(e);
47     }
48     priority_queue<P, vector<P>, greater<P> >myQueue;
49     myQueue.push(P(0, 1));
50     for (int i = 1; i <= n; i++)
51     {
52         d[i] = INF;
53     }
54     d[1] = 0;
55     while (!myQueue.empty())
56     {
57         P temp = myQueue.top();
58         myQueue.pop();
59         int v = temp.second;
60         if (d[v] < temp.first)
61             continue;
62         for (int i = 0; i < G[v].size(); i++)
63         {
64             edge e = G[v][i];
65             if (d[e.to] > d[v] + e.cost)
66             {
67                 d[e.to] = d[v] + e.cost;
68                 myQueue.push(P(d[e.to], e.to));
69             }
70         }
71     }
72     cout << d[n] << endl;
73     return 0;
74 }