hdu 4123 樹形DP+RMQ

life4711發表於2015-07-17
MQ

http://acm.hdu.edu.cn/showproblem.php?pid=4123

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
1
2
3
4
5
0 0





 



Sample Output




1
3
3
3
5




/**
hdu 4123 樹形DP+RMQ
題目大意：給定一棵樹，每一個點都從當前位置走到距離最遠的位置，1~n的連續區間中最大並且走的最遠距離差值不超過Q的區間右多大
解題思路：兩遍dfs求出在樹中到當前點的最長距離：dfs1求出以當前節點為根節點的子樹中到該節點的最長距離和次長距離，dfs2將上一步求出的
          最長距離和經過根節點的最長距離比較取最大。利用RMQ查詢尋找最大區間
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int N=50050;
int head[N],ip;
struct note
{
    int v,w,next;
}edge[N*2];

void init()
{
    memset(head,-1,sizeof(head));
    ip=0;
}

void addedge(int u,int v,int w)
{
    edge[ip].v=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++;
}

int maxn[N],smaxn[N],maxid[N],smaxid[N];

void dfs1(int u,int pre)
{
    maxn[u]=smaxn[u]=maxid[u]=smaxid[u]=0;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(pre==v)continue;
        dfs1(v,u);
        if(maxn[v]+edge[i].w>smaxn[u])
        {
            smaxid[u]=v;
            smaxn[u]=maxn[v]+edge[i].w;
            if(maxn[u]<smaxn[u])
            {
                swap(maxn[u],smaxn[u]);
                swap(maxid[u],smaxid[u]);
            }
        }
    }
}

void dfs2(int u,int pre)
{
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(pre==v)continue;
        if(maxid[u]==v)
        {
            if(smaxn[u]+edge[i].w>smaxn[v])
            {
                smaxn[v]=smaxn[u]+edge[i].w;
                smaxid[v]=u;
                if(maxn[v]<smaxn[v])
                {
                    swap(maxn[v],smaxn[v]);
                    swap(maxid[v],smaxid[v]);
                }
            }
        }
        else
        {
            if(maxn[u]+edge[i].w>smaxn[v])
            {
                smaxn[v]=maxn[u]+edge[i].w;
                smaxid[v]=u;
                if(maxn[v]<smaxn[v])
                {
                    swap(maxn[v],smaxn[v]);
                    swap(maxid[v],smaxid[v]);
                }
            }
        }
        dfs2(v,u);
    }
}

int a[N],n,m;
int dp1[N][30];
int dp2[N][30];

void RMQ_init(int n)
{
    for(int i=1;i<=n;i++)
    {
        dp1[i][0]=a[i];
        dp2[i][0]=a[i];
    }
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i+(1<<j)-1<=n;i++)///白書上的模板，次行稍作改動，否則dp陣列要擴大一倍防止RE
        {
            dp1[i][j]=max(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
            dp2[i][j]=min(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
        }
    }
}

int rmq(int x,int y)
{
    int k=0;
    while((1<<(k+1))<=y-x+1)k++;
    return max(dp1[x][k],dp1[y-(1<<k)+1][k])-min(dp2[x][k],dp2[y-(1<<k)+1][k]);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)break;
        init();
        for(int i=1;i<n;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        dfs1(1,-1);
        dfs2(1,-1);
        for(int i=1;i<=n;i++)
        {
            a[i]=maxn[i];
        }
        RMQ_init(n);
        while(m--)
        {
            int Q;
            scanf("%d",&Q);
            int ans=0;
            int id=1;
            for(int i=1;i<=n;i++)
            {
                while(id<=i&&rmq(id,i)>Q)id++;
                ans=max(ans,i-id+1);
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}




            




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