HDU 5438 Ponds dfs模拟

2015 ACM/ICPC Asia Regional Changchun Online

题意：n个池塘，删掉度数小于2的池塘，输出池塘数为奇数的连通块的池塘容量之和.

思路：两个dfs模拟就行了

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <fstream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <deque>
 7 #include <vector>
 8 #include <queue>
 9 #include <string>
10 #include <cstring>
11 #include <map>
12 #include <stack>
13 #include <set>
14 #define LL long long
15 #define eps 1e-8
16 #define INF 0x3f3f3f3f
17 #define MAXN 10005
18 using namespace std;
19 vector<int> G[MAXN];
20 struct Node{
21     int pos, du;
22     Node(int pos = 0, int du = 0):pos(pos), du(du){};
23 };
24 bool compare(Node x, Node y){
25     return x.du < y.du;
26 }
27 int find(int x){
28     if(father[x] == x) return x;
29     father[x] = find(father[x]);
30     return father[x];
31 }
32 Node node[MAXN];
33 int main()
34 {
35 #ifndef ONLINE_JUDGE
36     freopen("in.txt", "r", stdin);
37     //freopen("out.txt", "w", stdout);
38 #endif // OPEN_FILE
39     int T;
40     int p, m;
41     while (T--){
42         scanf("%d%d", &p, &m);
43         for (int i = 1; i <= p; i++){
44             scanf("%d", &v[i]);
45         }
46         int x, y;
47         for (int i = 1; i <= m; i++){
48             scanf("%d%d", &x, &y);
49             du[x]++;
50             du[y]++;
51             G[x].push_back(y);
52             G[y].push_back(x);
53         }
54         for(int i = 1; i <= n; i++){
55             node[i] = Node(i, du[i]);
56         }
57         sort(node + 1, node + n + 1, compare);
58         for(int i = 1; i <= n; i++){
59             int x = node[i].pos;
60             if(du[x] > 1) continue;
61             rm[x] = true;
62             for(int i = 0; i < G[x].size(); i++){
63                 du[G[x][i]]--;
64             }
65         }
66         for(int i = 1; i <= n; i++){
67             if(rm[i]) continue;
68             x = find(i);
69             for(int j = 0; j < G[i].size(); j++){
70                 int y = G[x][j];
71                 if(rm[y]) continue;
72                 y = find(y);
73                 if(x == y) continue;
74                 father[x] = y;
75             }
76         }
77         memset(cnt, 0, sizeof(cnt));
78         for(int i = 1; i <= n; i++){
79             cnt[father[i]]++;
80         }
81         for(int i = 1; i <= n; i++){
82             if(cnt[father[i]] & 1){
83                 ans += v[i];
84             }
85         }
86         printf("%d\n", ans);
87     }
88 }