hdu 5102(队列+节点扩展)

The K-th Distance

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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)

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

用一个队列维护即可，先将所有边入队，然后不停的用边的一个端点向外扩张，每扩张一次，路径长度就+1

由于1->2和2->1会重复算，所以K要扩大一倍，那么最后的结果肯定会多算一倍，除以2即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define N 100007

struct node
{
    int u,v,d;
    node(int _u,int _v,int _d):u(_u),v(_v),d(_d){}
    node(){}
};
struct Edge{
    int v,next;
}G[2*N];
int ans,k,tot,head[N];
queue<node> q;

void addedge(int u,int v)
{
    G[tot].v = v;
    G[tot].next = head[u];
    head[u] = tot++;
}

void bfs()
{
    int cnt = 0;
    while(!q.empty())
    {
        node tmp = q.front();
        q.pop();
        int u = tmp.u, v = tmp.v, d = tmp.d;
        if(cnt >= k) break;
        for(int i = head[u]; i != -1; i = G[i].next)
        {
            int vv = G[i].v;
            if(vv != v)
            {
                ans += d+1;
                cnt++;
                q.push(node(vv,u,d+1));
            }
            if(cnt >= k) break;
        }
    }
}

int main()
{
    int t,n,u,v,i;
    scanf("%d",&t);
    while(t--)
    {
        while(!q.empty()) q.pop();
        scanf("%d%d",&n,&k);
        tot = 0;
        memset(head,-1,sizeof(head));
        for(i = 1; i <= n; i++) q.push(node(i,i,0));
        for(i = 1; i < n; i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        k *= 2;
        ans = 0;
        bfs();
        cout<<ans/2<<endl;
    }
    return 0;
}