hdu5438 Ponds-优快云博客

本文介绍了一种算法，用于计算在特定条件下剩余池塘的价值总和。该算法首先去除度数小于2的节点，然后计算剩余强连通组件中奇数个节点的价值总和。

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Ponds

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 985 Accepted Submission(s): 328

Problem Description

Betty owns a lot of ponds, some of them are connected with other ponds by pipes, and there will not be more than one pipe between two ponds. Each pond has a value

Input

The first line of input will contain a number

Output

For each test case, output the sum of the value of all connected components consisting of odd number of ponds after removing all the ponds connected with less than two pipes.

Sample Input

1
7 7
1 2 3 4 5 6 7
1 4
1 5
4 5
2 3
2 6
3 6
2 7

Sample Output

Source

2015 ACM/ICPC Asia Regional Changchun Online

先说题意吧：

有n个池塘通过m个水管连通（2个池塘之间只有一个水管）现在要以走一些池塘（条件是这个池塘只连接了一个水管）一直移到不能够再移除为止

求在剩下的池塘中由奇数个池塘组成的池塘的价值的和

思路：去掉这些度数小于2的点后剩下的就都是强连通图了

找度数大于1的数量为奇数的强联通分量的和（奇数！！！这道题没有自环和重边）

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <iostream>
 5 #include <algorithm>
 6 using namespace std;
 7 const int oo = 0x3f3f3f3f;
 8 const int N = 100005;
 9 typedef long long LL;
10 struct da
11 {
12     int u, v, next;
13 }as[N*4];
14 int head[N], cnt, n, m, vis[N], in[N];
15 /**< in记录每个点的度数  vis标记这个点是否处理过 val记录每个池塘的价值 */
16 LL val[N];
17 void add(int u, int v)
18 {
19     as[cnt].v = v;
20     as[cnt].next = head[u];
21     head[u] = cnt++;
22 }
23 void init()
24 {
25     memset(head, -1, sizeof(head));
26     memset(in, 0, sizeof(in));
27     cnt = 0;
28 }
29 void dfs(int u)/**< 将度数小于2的点以及这个点能到达的边的度数都减去1 */
30 {
31     vis[u] = 1;in[u]--;
32     for(int i = head[u]; i != -1; i = as[i].next)
33     {
34         int v = as[i].v;
35         if(vis[v]) continue;/**< 因为是无向图 有1-2 就有2-1 避免通一条边多次处理 */
36         in[v]--;
37         if(in[v] < 2)/**< 如果这个点的度数小于2 处理这个点的连边*/
38            dfs(v);
39     }
40 }
41 void dfs1(int u, LL &num, LL &sum)/**< 强连通的池塘的价值和 */
42 {
43     vis[u] = 1;
44     num++; sum += val[u];
45     for(int i = head[u]; i != -1; i = as[i].next)
46     {
47         int v = as[i].v;
48         if(vis[v]) continue;
49         if(in[v] < 2) continue;
50         dfs1(v, num, sum);
51     }
52 }
53 LL solve()
54 {
55     int i;
56     memset(vis, 0, sizeof(vis));
57     for(i = 1; i <= n; i++)
58     {
59         if(vis[i] == 0 && in[i] < 2)/**< 如果这个点没有遍历过 并且度数小于2 对他的连边处理 */
60             dfs(i);
61     }
62     LL sum=0, ans = 0, QQ = 0;
63     memset(vis, 0, sizeof(vis));
64     for(i = 1; i <= n; i++)
65     {
66         QQ = 0, sum = 0;
67         if(vis[i] == 0 && in[i] >= 2)
68         {
69            dfs1(i, sum, QQ);
70            if(sum%2)ans += QQ;/**< 找出池塘个数为奇数的和 */
71         }
72     }
73     return ans;
74 }
75 int main()
76 {
77     int T, i;
78     scanf("%d", &T);
79     while(T--)
80     {
81         init();
82         scanf("%d %d", &n, &m);
83         for(i = 1; i <= n; i++)
84             scanf("%I64d", &val[i]);
85         while(m--)
86         {
87             int a, b;
88             scanf("%d %d", &a, &b);
89             if(a == b)continue;
90             add(a, b);
91             add(b, a);
92             in[a]++;
93             in[b]++;
94         }
95         LL ans = solve();
96         printf("%I64d\n", ans);
97     }
98     return 0;
99 }