uva12118

一开始以为直接算联通块个数就行了
后来发现还得分联通块里的奇点。。。
还要注意m = 0的情况...
 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<cstring>
 6 #include<string>
 7 
 8 using namespace std;
 9 
10 void setIO(const string& a) {
11     freopen((a+".in").c_str(), "r", stdin);
12     freopen((a+".out").c_str(), "w", stdout);
13 }
14 
15 const int N = 1000 + 10;
16 
17 #include<vector>
18 vector<int> G[N];
19 bool vis[N];
20 int du[N];
21 
22 int dfs(int u) {
23     vis[u] = 1;
24     int res = G[u].size() & 1;
25     for(unsigned i = 0; i < G[u].size(); i++) {
26         int v = G[u][i];
27         if(!vis[v]) res += dfs(v);
28     }
29     return res;
30 }
31     
32 
33 int main() {
34 #ifdef DEBUG
35     freopen("in.txt", "r", stdin);
36     freopen("out.txt", "w", stdout);
37 #endif
38     
39     int n, m, T, cas = 0;
40     while(scanf("%d%d%d", &n, &m, &T) && n) {
41         for(int i = 1; i <= n; i++) {
42             G[i].clear();
43             du[i] = 0;
44             vis[i] = 0;
45         }
46         for(int i = 1; i <= m; i++) {
47             int u, v;
48             scanf("%d%d", &u, &v);
49             G[u].push_back(v);
50             G[v].push_back(u);
51             du[u]++, du[v]++;
52         }
53         int c1 = 0, c2 = 0;
54         for(int i = 1; i <= n; i++) if(!vis[i] && du[i]) {
55             c1++; c2 += max((dfs(i) - 2) / 2, 0);
56         }
57         
58         printf("Case %d: %d
", ++cas, T * (max(c1 - 1, 0) + c2 + m));
59     }
60     
61     return 0;
62 }