hdu 5102 The K-th Distance

The K-th Distance

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 27    Accepted Submission(s): 5

Problem Description

Given a tree, which has n node in total. Define the distance between two node u and v is the number of edge on their unique route. So we can have n(n-1)/2 numbers for all the distance, then sort the numbers in ascending order. The task is to output the sum of the first K numbers.

 

 

Input

There are several cases, first is the number of cases T. (There are most twenty cases).
For each case, the first line contain two integer n and K (2≤n≤100000,0≤K≤min(n(n−1)/2,106) ). In following there are n-1 lines. Each line has two integer u , v. indicate that there is an edge between node u and v.

 

 

Output

For each case output the answer.

 

 

Sample Input

2 3 3 1 2 2 3 5 7 1 2 1 3 2 4 2 5

 

 

Sample Output

4 10

 

 

Source

BestCoder Round #17

 

思路： 因为k <= 10^6 ，假设是一条链，n = 10^5 ，长度为1的有n-1，长度为2的有n-2....

那么长度不会超过20，这个我们就可以只计算长度为k/n+70的，

直接树分治搞就好了 

 

#pragma comment(linker,"/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<set>#include<stack>#include<map>#include<ctime>#include<bitset>#define LL long long#define maxn 100010#define INF 0x3f3f3f3fusing namespace std;int head[maxn],to[maxn*2],next1[maxn*2] ;int top,cnt[maxn],len ;int f[maxn];bool vi[maxn] ;void Unit(int u,int v){    next1[top] = head[u] ;to[top]=v;    head[u]=top++;}void find1( int u ,int fa ){    f[u] = 0 ;    cnt[u] = 1 ;    int v ;    for( int i = head[u] ; i != -1; i = next1[i])    {         v = to[i] ;        if(v==fa||vi[v]) continue ;        find1(v,u) ;        cnt[u] += cnt[v] ;        f[u] = max(cnt[v],f[u]) ;    }}int minn,k,n,MAX ;void find_root1( int u ,int fa,int &root,int sum){    int tmp = max(sum-cnt[u],f[u]) ;    if(tmp < minn)    {        minn = tmp ;        root = u ;    }    int v;    for( int i = head[u] ; i != -1; i = next1[i])    {        v = to[i] ;        if(v==fa||vi[v]) continue ;        find_root1(v,u,root,sum) ;    }}int get_root( int u ){    find1(u,-1) ;    int sum = cnt[u] ;    int root = u ;    minn = n ;    find_root1(u,-1,root,sum) ;    return root ;}int que[maxn],tt;int num[maxn],ans[maxn] ;vector<int>vec;void find(int len){    for(int i = 1 ; i <= MAX ;i++)if(i+len<=MAX)    {        ans[i+len] += num[i] ;    }}void dfs(int u,int fa,int len1){    int v ;    ans[len1]++;    vec.push_back(len1);    find(len1) ;    for( int i = head[u] ; i != -1; i = next1[i])    {        v = to[i] ;        if(v==fa||vi[v]||len1+1>MAX) continue ;        dfs(v,u,len1+1) ;    }}void count(int u){    int v ,j ;    //memset(num,0,sizeof(num));    tt=0;    for(int i = head[u] ; i != -1; i = next1[i])    {        v = to[i] ;        if(vi[v]) continue ;        len=0;        vec.clear();        dfs(v,u,1);        for( j = 0 ; j < vec.size();j++)        {            if(!num[vec[j]])que[tt++]=vec[j];             num[vec[j]]++;        }    }    for(int i = 0 ; i < tt;i++)        num[que[i]]=0;}void solve(int u){    int root=get_root(u);    vi[root]=true;    count(root);    for(int i = head[root] ; i != -1; i = next1[i])    {        int v = to[i] ;        if(vi[v]) continue ;        solve(v) ;    }}void init(){    len = 0 ;    memset(vi,0,sizeof(vi)) ;    top=0;    memset(head,-1,sizeof(head)) ;    memset(num,0,sizeof(num)) ;    memset(ans,0,sizeof(ans)) ;}int main(){    int i ,j ,m ;    int T,u,v;    // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    cin >> T ;    while(T--)    {        scanf("%d%d",&n,&k) ;        MAX=min(k/n+70,k);        init();        for( i = 1 ; i < n ;i++)        {            scanf("%d%d",&u,&v);            Unit(u,v) ;            Unit(v,u) ;        }        solve(1);        LL sum=0;        for( i = 1 ; i <= MAX ;i++)        {            if(ans[i]>=k){                sum += (LL)k*i ;                break;            }            k -= ans[i] ;            sum += (LL)ans[i]*i ;        }        printf("%I64d\n",sum) ;    }    return 0 ;}