Y
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2284 Accepted Submission(s): 605
Problem Description

Sample Input
4 1 2 1 3 1 4
Sample Output
1Hint1. The only set is {2,3,4}. 2. Please use #pragma comment(linker, "/STACK:16777216")
Source
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题目大意:从一颗树上找3个节点,这三个节点之间没有一条路相连,
思路:用总的情况的个数减去满足三个节点之间有路的情况个数,满足的个数求法为选i节点,再选i的子树一个节点,再选树的其他部分一个节点,会重复,求出来除2
ac代码
#include<stdio.h>
#include<string.h>
#pragma comment(linker, "/STACK:16777216")
int son[100010],head[100010],dig[100010],cnt,f[100010];
struct s
{
int u,v,next;
}edge[200020];
void add(int u,int v)
{
edge[cnt].u=u;
edge[cnt].v=v;
//edge[cnt].w=w;
edge[cnt].next=head[u];
head[u]=cnt++;
}
int dfs(int u,int pre)
{
son[u]=1;
f[u]=pre;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v==pre)
continue;
son[u]+=dfs(v,u);
}
return son[u];
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int i,j;
memset(head,-1,sizeof(head));
cnt=0;
for(i=1;i<=n-1;i++)
{
int u,v;
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
dfs(1,-1);
__int64 sum=0,a,b;
for(i=1;i<=n;i++)
{
__int64 temp=0;
for(j=head[i];j!=-1;j=edge[j].next)
{
int v=edge[j].v;
if(v==f[i])
a=n-son[i];
else
a=son[v];
b=n-a-1;
temp+=a*b;
}
sum+=temp/2;
}
__int64 ans=(__int64)n*(n-1)*(n-2)/6-sum;
printf("%I64d\n",ans);
}
}