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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 60    Accepted Submission(s): 20

Problem Description  

Sample Input 4 1 2 1 3 1 4  

Sample Output 1
Hint 1. The only set is {2,3,4}. 2. Please use #pragma comment(linker, “/STACK:16777216”)  

Source
2013 Multi-University Training Contest 10  

Recommend zhuyuanchen520  

树形DP统计一遍就可以了。

 1 /* ***********************************************
 2 Author        :kuangbin
 3 Created Time  :2013/8/22 12:27:59
 4 File Name     :F:\2013ACM练习\2013多校10\1010.cpp
 5 ************************************************ */
 6 #pragma comment(linker, "/STACK:1024000000,1024000000")
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 
21 const int MAXN = 100010;
22 struct Edge
23 {
24     int to,next;
25 }edge[MAXN*2];
26 int head[MAXN],tot;
27 void init()
28 {
29     memset(head,-1,sizeof(head));
30     tot = 0;
31 }
32 void addedge(int u,int v)
33 {
34     edge[tot].to = v;
35     edge[tot].next = head[u];
36     head[u] = tot++;
37 }
38 
39 
40 int num[MAXN];
41 int n;
42 long long ans;
43 void dfs(int u,int pre)
44 {
45     num[u] = 1;
46     int tmp = 0;
47     for(int i = head[u];i!= -1;i = edge[i].next)
48     {
49         int v = edge[i].to;
50         if(v == pre)continue;
51         dfs(v,u);
52         ans += (long long)tmp*num[v];
53         num[u] += num[v];
54         tmp += num[v];
55     }
56     ans += (long long)tmp*(n-num[u]);
57 }
58 
59 
60 int main()
61 {
62     //freopen("in.txt","r",stdin);
63     //freopen("out.txt","w",stdout);
64     int u,v;
65     while(scanf("%d",&n) == 1)
66     {
67         init();
68         for(int i = 1;i < n;i++)
69         {
70             scanf("%d%d",&u,&v);
71             addedge(u,v);
72             addedge(v,u);
73         }
74         ans = 0;
75         dfs(1,-1);
76         long long tot = (long long)n*(n-1)*(n-2)/6;
77         printf("%I64d\n",tot-ans);
78     }
79     return 0;
80 }