HDU 4705

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：

# include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

const int MAX = 100010;
typedef long long int ll;
struct node
{
    int to;
    int next;
}tree[MAX * 2];
int head[MAX], num[MAX];
int tol = 0;
int n;
ll sum;
void add(int a, int b)
{
    tree[tol].to = b;
    tree[tol].next = head[a];
    head[a] = tol++;
}

void dfs(int root, int f)
{
    int tep = 0;
    for(int i = head[root]; i != -1; i = tree[i].next)
    {
        int son = tree[i].to;
        if(son == f)
            continue;
        dfs(son, root);
        
        num[root] += num[son];
        tep += num[son];
        sum += (ll) (n - tep - 1) * num[son];
    }
}

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        tol = 0;
        sum = 0;
        memset(head, -1, sizeof(head));
        for(int i = 0; i < MAX; i++)
            num[i] = 1;
        int a, b;
        for(int i = 1; i < n; i++)
        {
            scanf("%d%d", &a, &b);
            add(a, b);
            add(b, a);
        }
        dfs(1, -1);
        ll all = (ll)n * (n - 1) * (n - 2) / 6;
        
        printf("%I64d
", all - sum);
    }
    return 0;
}