POJ 3177 Redundant Paths(Tarjan)

题目链接

看这个题解：http://blog.csdn.net/zhang20072844/article/details/8145588

对于Tarjan还是不太理解。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cstdlib>
 5 using namespace std;
 6 #define N 10001
 7 #define M 50001
 8 struct node
 9 {
10     int u,v,next;
11 } edge[M+10];
12 int t,top,scnt;
13 int DFN[N+1],Low[N+1],first[N+1];
14 int cnt[N+1];
15 bool mat[5001][5001];
16 void CL()
17 {
18     t = 0;
19     t = scnt = top = 0;
20     memset(first,-1,sizeof(first));
21     memset(DFN,0,sizeof(DFN));
22     memset(mat,0,sizeof(mat));
23 }
24 void add(int u,int v)
25 {
26     edge[t].u = u;
27     edge[t].v = v;
28     edge[t].next = first[u];
29     first[u] = t ++;
30 }
31 void Tarjan(int u,int father)
32 {
33     int v,i;
34     DFN[u] = Low[u] = ++ t;
35     for(i = first[u]; i != -1; i = edge[i].next)
36     {
37         v = edge[i].v;
38         if(!DFN[v])
39         {
40             Tarjan(v,u);
41             if(Low[u] > Low[v])
42             {
43                 Low[u] = Low[v];
44             }
45         }
46         else if(v != father&&Low[u] > DFN[v])
47         {
48             Low[u] = DFN[v];
49         }
50     }
51 }
52 int main()
53 {
54     int n,m,sv,ev,i,j,v,ans;
55     while(scanf("%d%d",&n,&m)!=EOF)
56     {
57         CL();
58         memset(cnt,0,sizeof(cnt));
59         for(i = 1; i <= m; i ++)
60         {
61             scanf("%d%d",&sv,&ev);
62             if(!mat[sv][ev])
63             {
64                 add(sv,ev);
65                 add(ev,sv);
66                 mat[sv][ev] = mat[ev][sv] = true;
67             }
68         }
69         t = 0;
70         Tarjan(1,1);
71         for(i = 1;i <= n;i ++)
72         {
73             for(j = first[i];j != -1;j = edge[j].next)
74             {
75                 v = edge[j].v;
76                 if(Low[v] != Low[i])
77                 cnt[Low[i]] ++;
78             }
79         }
80         ans = 0;
81         for(i = 1;i <= n;i ++)
82         {
83             if(cnt[i] == 1)
84             ans ++;
85         }
86         printf("%d\n",(ans+1)/2);
87     }
88     return 0;
89 }

 

posted @ 2013-04-16 19:52  Naix_x  阅读(136)  评论(0)  编辑  收藏  举报