Bob’s Race

Time Limit: 5000/2000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads in his village. Each road connects two houses, and all houses are connected together. To make the race more interesting, he requires that every participant must start from a different house and run AS FAR AS POSSIBLE without passing a road more than once. The distance difference between the one who runs the longest distance and the one who runs the shortest distance is called “race difference” by Bob. Bob does not want the “race difference”to be more than Q. The houses are numbered from 1 to N. Bob wants that the No. of all starting house must be consecutive. He is now asking you for help. He wants to know the maximum number of starting houses he can choose, by other words, the maximum number of people who can take part in his race.

Input

There are several test cases.
The first line of each test case contains two integers N and M. N is the number of houses, M is the number of queries.
The following N-1 lines, each contains three integers, x, y and z, indicating that there is a road of length z connecting house x and house y.
The following M lines are the queries. Each line contains an integer Q, asking that at most how many people can take part in Bob’s race according to the above mentioned rules and under the condition that the“race difference”is no more than Q.

The input ends with N = 0 and M = 0.

(N<=50000 M<=500 1<=x,y<=N 0<=z<=5000 Q<=10000000)

Output

For each test case, you should output the answer in a line for each query.

Sample Input

5 5

1 2 3

2 3 4

4 5 3

3 4 2

0 0

Sample Output

题意：

给一棵树,从每个点出发,这点的权值为这点能到的最远的距离;有m个询问,问连续的ans个点的权值<=q,(ans为要求的最大值);

思路：

先3遍的bfs找出每个点的权值,然后用ST算法求出RMQ;然后用尺取法来找出答案,也可以用单调队列搞定;

AC代码：

/*4123    717MS    13772K    3360 B    G++    2014300227*/
#include <bits/stdc++.h>
#include <cmath>
using namespace std;
const int N=50004;
int n,m,u,v,w,cnt,head[N];
struct Edge
{
    int to,next,val;
};
Edge edge[2*N];
void add_edge(int s,int e,int va)
{
    edge[cnt].to=e;
    edge[cnt].next=head[s];
    edge[cnt].val=va;
    head[s]=cnt++;
}
int dis1[N],dis2[N],dis[N],pin[N][25],pax[N][25],vis[N],f[N];
queue<int>qu;
int bfs()
{
    memset(vis,0,sizeof(vis));
    //memset(dis1,0,sizeof(dis1));
    qu.push(1);
    vis[1]=1;
    dis1[1]=0;
    while(!qu.empty())
    {
        int fr=qu.front();
        qu.pop();
        for(int i=head[fr];i!=-1;i=edge[i].next)
        {
            int fx=edge[i].to;
            if(!vis[fx])
            {
                dis1[fx]=dis1[fr]+edge[i].val;
                vis[fx]=1;
                qu.push(fx);
            }
        }
    }
}
int bfs1()
{
    int ans=1;
    for(int i=1;i<=n;i++)
    {
        if(dis1[i]>=dis1[ans])
        {
            ans=i;
        }
    }
    //memset(dis1,0,sizeof(dis1));
    memset(vis,0,sizeof(vis));
    qu.push(ans);
    vis[ans]=1;
    dis1[ans]=0;
    while(!qu.empty())
    {
        int fr=qu.front();
        qu.pop();
        for(int i=head[fr];i!=-1;i=edge[i].next)
        {
            int fx=edge[i].to;
            if(!vis[fx])
            {
                dis1[fx]=dis1[fr]+edge[i].val;
                vis[fx]=1;
                qu.push(fx);
            }
        }
    }
}
int bfs2()
{
    //memset(dis2,0,sizeof(dis2));
    memset(vis,0,sizeof(vis));
    int ans=1;
    for(int i=1;i<=n;i++)
    {
        if(dis1[i]>=dis1[ans])
        {
            ans=i;
        }
    }
    qu.push(ans);
    vis[ans]=1;
    dis2[ans]=0;
    while(!qu.empty())
    {
        int fr=qu.front();
        qu.pop();
        for(int i=head[fr];i!=-1;i=edge[i].next)
        {
            int fx=edge[i].to;
            if(!vis[fx])
            {
                vis[fx]=1;
                dis2[fx]=dis2[fr]+edge[i].val;
                qu.push(fx);
            }
        }
    }
}
int rmq()
{

    for(int i=1;i<=n;i++)
    {
        f[i]=(int)(log(i*1.0)/log(2.0));
        dis[i]=max(dis1[i],dis2[i]);
       pax[i][0]=pin[i][0]=dis[i];

    }
    for(int j=1;j<=f[n];j++)
    {
        for(int i=1;i+(1<<(j-1))<=n+1;i++)
        {
            pin[i][j]=min(pin[i][j-1],pin[i+(1<<(j-1))][j-1]);
            pax[i][j]=max(pax[i][j-1],pax[i+(1<<(j-1))][j-1]);
        }
    }
}
int get_ans(int x)
{
    int l=1,r=1,ans=0,mmax,mmin;
    while(r<=n)
    {
        int temp=f[r-l+1];
        mmax=max(pax[l][temp],pax[r-(1<<temp)+1][temp]);
        mmin=min(pin[l][temp],pin[r-(1<<temp)+1][temp]);
        if(mmax-mmin<=x)ans=max(ans,r-l+1),r++;
        else l++;
    }
    return ans;
}
int main()
{
    while(1)
    {
        scanf("%d%d",&n,&m);
        if(n==0&&m==0)break;
        cnt=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<n;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add_edge(u,v,w);
            add_edge(v,u,w);
        }
        bfs();
        bfs1();
        bfs2();
        rmq();
        while(m--)
        {
            int q;
            scanf("%d",&q);
            printf("%d
",get_ans(q));
        }

    }
}