HDU 4705 Y

Y

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 433    Accepted Submission(s): 147

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

 

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zhuyuanchen520
 
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#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
#include <cmath>
#define N 100010
#pragma comment(linker, "/STACK:16777216")
__int64 two[N];
__int64 sum[N];
bool status[N];
int level[N];
struct num
{
    int e,next;
}a[2*N];
int b[N],Top;
int main()
{
    //freopen("data.in","r",stdin);
    void addeage(int x,int y);
    __int64 pre_deal(int k);
    __int64 deal(int k);
    __int64 get_two(int k);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        Top=0;
        memset(b,-1,sizeof(b));
        for(int i=1;i<=n-1;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            addeage(x,y);
            addeage(y,x);
        }
        memset(sum,0,sizeof(sum));
        memset(status,false,sizeof(status));
        memset(level,0,sizeof(level));
        pre_deal(1);
        memset(two,0,sizeof(two));
        get_two(1);
        memset(status,false,sizeof(status));
        __int64 res=deal(1);
        printf("%I64d
",res);
    }
    return 0;
}
void addeage(int x,int y)
{
    a[Top].e = y;
    a[Top].next = b[x];
    b[x]=Top++;
}
__int64 pre_deal(int k)
{
    status[k]=true;
    for(int i=b[k];i!=-1;i=a[i].next)
    {
        int x = a[i].e;
        if(!status[x])
        {
            level[x]=level[k]+1;
            sum[k]+=pre_deal(x);
        }
    }
    sum[k]+=1;
    return sum[k];
}
__int64 get_two(int k)
{
    __int64 s1;
    bool check=true;
    for(int i=b[k];i!=-1;i=a[i].next)
    {
        int y = a[i].e;
        if(level[y]!=level[k]+1)
        {
            continue;
        }
        if(check)
        {
            s1=sum[y];
            check=false;
            continue;
        }
        two[k]+=(s1*sum[y]);
        s1+=sum[y];
    }
    for(int i=b[k];i!=-1;i=a[i].next)
    {
        int y = a[i].e;
        if(level[y]!=level[k]+1)
        {
            continue;
        }
        two[k]+=get_two(y);
    }
    return two[k];
}
__int64 deal(int k)
{
    status[k]=true;
    __int64 s = 0;
    __int64 x2,three=0,s1,temp=0;
    int uv=0;
    bool check=true;
    for(int i=b[k];i!=-1;i=a[i].next)
    {
        int y = a[i].e;
        if(level[y]!=level[k]+1)
        {
            continue;
        }
        if(uv==0)
        {
            s1=sum[y];
            uv++;
            continue;
        }
        if(check)
        {
            temp+=(s1*sum[y]);
            s1+=sum[y];
            check=false;
            continue;
        }
        three+=(temp*sum[y]);
        temp+=(s1*sum[y]);
        s1+=sum[y];
    }
    s+=three;
    __int64 w=0;
    for(int i=b[k];i!=-1;i=a[i].next)
    {
        if(!status[a[i].e])
        {
            s+=deal(a[i].e);
            w+=(sum[a[i].e]);
            s+=two[a[i].e];
        }
    }
    __int64 ans=0;
    for(int i=b[k];i!=-1;i=a[i].next)
    {
        int y =a[i].e;
        __int64 res=0;
        if(level[y]==level[k]+1&&two[y]!=0)
        {
            res+=((w-sum[y])*two[y]);
        }
        ans+=res;
    }
    s+=ans;
    return s;
}