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) ). 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

23 31 22 35 71 21 32 42 5

 

Sample Output

410
解题思路：这道题一开始的想法就是bfs，因为bfs是按层次来扩展节点，可以算出每个节点到根节点的距离。但关键就是两点之间的距离如果没有横跨根节点，这里想了好久，一直没有办法解决。看了别人的思路，说实话很难想到，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 100007struct 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;}