Bob’s Race

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

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 

HINT

题意

一个城镇有N个住户，N-1条路连接两个住户，保证N个住户联通，M次询问，给定N条边的信息，包括连
接的住户序号以及路的长度。然后是M次询问，每次询问Q，要求找到最长的连续序号，使得Max(dis[i]) – Min(dis[i]) ≤
Q(l≤i≤r)，输出最大的r-l+1。dis[i]为从第i个住户出发，不重复走过路能移动的最远距离。

题解：

dis[i]一定是点到直径的某个点的距离，所以我们两次dfs求出直径，然后两次dfs求出距离就好了

至于第二问，对于每个询问，我们可以O(N)扫一遍就好了

代码

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<vector>
using namespace std;

vector<pair<int,int> > Q[50005];
int d[50005];
int n,m;
int left1,l_num;
int Right,r_num;
int dpMax[50005][20], dpMin[50005][20];
void dfs1(int x,int pre,int len,int type)
{
    if(type == 1 && l_num < len)
    {
        l_num = len;
        left1 = x;
    }
    if(type == 2 && r_num < len)
    {
        r_num = len;
        Right = x;
    }
    for(int i=0;i<Q[x].size();i++)
    {
        pair<int,int> K = Q[x][i];
        if(K.first == pre)continue;
        dfs1(K.first,x,len + K.second,type);
    }
}
void dfs2(int x,int pre,int len)
{
    d[x]=max(d[x],len);
    for(int i=0;i<Q[x].size();i++)
    {
        pair<int,int> K = Q[x][i];
        if(K.first == pre)continue;
        dfs2(K.first,x,len + K.second);
    }
}
void rmq_init() {
    for (int i = 1; i <= n; i++)
        dpMax[i][0] = dpMin[i][0] = d[i];

    for (int k = 1; (1<<k) <= n; k++) {
        for (int i = 1; i + (1<<k) - 1 <= n; i++) {
            dpMax[i][k] = max(dpMax[i][k-1], dpMax[i+(1<<(k-1))][k-1]);
            dpMin[i][k] = min(dpMin[i][k-1], dpMin[i+(1<<(k-1))][k-1]);
        }
    }
}

int rmq_query(int l, int r) {
    int k = 0;
    while ((1<<(k+1)) <= r - l + 1) k++;
    return max(dpMax[l][k], dpMax[r-(1<<k)+1][k]) - min(dpMin[l][k], dpMin[r-(1<<k)+1][k]);
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0)break;
        memset(d,0,sizeof(d));
        memset(dpMax,0,sizeof(dpMax));
        memset(dpMin,0,sizeof(dpMin));
        for(int i=1;i<=n;i++)
            Q[i].clear();
        for(int i=1;i<n;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            Q[x].push_back(make_pair(y,z));
            Q[y].push_back(make_pair(x,z));
        }
        left1 = l_num = Right = r_num = 0;

        dfs1(1,-1,0,1);
        dfs1(left1,-1,0,2);
        dfs2(left1,-1,0);
        dfs2(Right,-1,0);
        rmq_init();
        while (m--) {
            int ans = 0, mv = 1;
            int QQ;
            scanf("%d", &QQ);
            for (int i = 1; i <= n; i++) {
                while (mv <= i && rmq_query(mv , i) > QQ) mv++;
                ans = max(ans, i - mv + 1);
            }
            printf("%d\n", ans);
        }

    }
}