HDU 4705 - Y

Posted 2020-08-04 stort

Y

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2575    Accepted Submission(s): 723

Problem Description

 

 

Sample Input

 

4

1 2

1 3

1 4

 

Sample Output

 

1

 

题意：

　　给你一棵树，求三个点不连续的个数

思路:

　　总的减去，3个在一条路径上的，以中间的数为 根，那么对于这个点就是，(n - his son - his) * his son

AC代码：

 1 # include <bits/stdc++.h>
 2 #pragma comment(linker, "/STACK:1024000000,1024000000")
 3 using namespace std;
 4 
 5 const int MAX = 100010;
 6 typedef long long int ll;
 7 struct node
 8 {
 9     int to;
10     int next;
11 }tree[MAX * 2];
12 int head[MAX], num[MAX];
13 int tol = 0;
14 int n;
15 ll sum;
16 void add(int a, int b)
17 {
18     tree[tol].to = b;
19     tree[tol].next = head[a];
20     head[a] = tol++;
21 }
22 
23 void dfs(int root, int f)
24 {
25     int tep = 0;
26     for(int i = head[root]; i != -1; i = tree[i].next)
27     {
28         int son = tree[i].to;
29         if(son == f)
30             continue;
31         dfs(son, root);
32         
33         num[root] += num[son];
34         tep += num[son];
35         sum += (ll) (n - tep - 1) * num[son];
36     }
37 }
38 
39 int main()
40 {
41     while(scanf("%d", &n) != EOF)
42     {
43         tol = 0;
44         sum = 0;
45         memset(head, -1, sizeof(head));
46         for(int i = 0; i < MAX; i++)
47             num[i] = 1;
48         int a, b;
49         for(int i = 1; i < n; i++)
50         {
51             scanf("%d%d", &a, &b);
52             add(a, b);
53             add(b, a);
54         }
55         dfs(1, -1);
56         ll all = (ll)n * (n - 1) * (n - 2) / 6;
57         
58         printf("%I64d\\n", all - sum);
59     }
60     return 0;
61 }

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