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POJ 3268 Silver Cow Party

时间:2018-06-25 23:01:05      阅读:325      评论:0      收藏:0      [点我收藏+]

标签:AC   pac   inf   +=   code   ret   clu   graph   art   

题意:给定顶点个数,边的个数,目的地,求顶点到边的往返最大距离

解题思路:目的地到顶点的是标准的Dijkstra,顶点到目的地的只需要把矩阵转置一下就OKay了

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define MAXVERTEXNUM 1005
 5 #define INF 1000000
 6 using namespace std;
 7 
 8 int Nv, Ne, Des;
 9 int MGraph[MAXVERTEXNUM][MAXVERTEXNUM];
10 int dist1[MAXVERTEXNUM], dist2[MAXVERTEXNUM];
11 
12 void Dijkstra()
13 {
14     int collected[MAXVERTEXNUM];
15     memset(collected, 0, sizeof(collected));
16     collected[Des] = 1;
17     dist1[Des] = 0;
18 
19     for (int i = 1; i <= Nv; ++i)
20     {
21         if (MGraph[Des][i] != INF)
22             dist1[i] = MGraph[Des][i];
23         else
24             dist1[i] = INF;
25     }
26 
27     for (int i = 2; i <= Nv; ++i)
28     {
29         int minDis = INF, num;
30         for (int j = 1; j <= Nv; ++j)
31             if (!collected[j] && minDis > dist1[j])
32             {
33                 minDis = dist1[j];
34                 num = j;
35             }
36 
37         collected[num] = 1;
38         dist1[num] = minDis;
39 
40         for (int j = 1; j <= Nv; ++j)
41             if (!collected[j] && dist1[j] > dist1[num] + MGraph[num][j])
42                 dist1[j] = dist1[num] + MGraph[num][j];
43     }
44 }
45 
46 void T()
47 {
48     for (int i = 1; i <= Nv; ++i)
49         for (int j = 1; j <= i; ++j)
50             swap(MGraph[i][j], MGraph[j][i]);
51 }
52 
53 int main()
54 {
55     ios::sync_with_stdio(false);
56 
57     cin >> Nv >> Ne >> Des;
58 
59     for (int i = 1; i <= Nv; ++i)
60         for (int j = 1; j <= Nv; ++j)
61             if (i != j)
62                 MGraph[i][j] = INF;
63             else
64                 MGraph[i][j] = 0;
65 
66     for (int i = 1; i <= Ne; ++i)
67     {
68         int V1, V2, Edge;
69         cin >> V1 >> V2 >> Edge;
70         MGraph[V1][V2] = Edge;
71     }
72 
73     Dijkstra();
74     for (int i = 1; i <= Nv; ++i)
75         dist2[i] = dist1[i];
76     T();
77     Dijkstra();
78 
79     int maxDis = 0;
80     for (int i = 1; i <= Nv; ++i)
81         if (i != Des)
82         {
83             int temp = 0;
84             temp += dist1[i];
85             temp += dist2[i];
86             if (temp > maxDis)
87                 maxDis = temp;
88         }
89 
90     cout << maxDis << endl;
91 
92     return 0;
93 }

 

POJ 3268 Silver Cow Party

标签:AC   pac   inf   +=   code   ret   clu   graph   art   

原文地址:https://www.cnblogs.com/ducklu/p/9226524.html

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