Bob’s Race
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1966 Accepted Submission(s): 607
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.
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 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
Source
Recommend
题意：
给你一棵树。定义一个结点到其它结点的最大长度为该点的路径长度。现给你一个q值。问你最多有多少个标号连续的结点满足。其中路径的最大值和最小值之差不超过q。
思路：
先树形DP求出一个结点的路径值。然后RMQ维护区间的最大最小值。然后l,r指针扫描区间就行了。比赛时就该写的。可是因为卡题了。所以就没写的机会了。
详细见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
struct node
{
int v,md,len;
node *next;
}edge[maxn<<1],*head[maxn];
int dp[maxn],cnt,n,m;
int vis[maxn],lg[maxn],rmqb[25][maxn],rmqs[25][maxn];
void init()
{
cnt=0;
memset(vis,0,sizeof vis);
memset(head,0,sizeof head);
}
void pre_rmq()
{
int i;
lg[0]=-1;
for(i=1;i<maxn;i++)
lg[i]=lg[i>>1]+1;
}
void rmq_init()
{
int i,j;
for(i=1;i<=n;i++)
rmqb[0][i]=rmqs[0][i]=dp[i];
for(i=1;i<=lg[n];i++)
for(j=1;j+(1<<i)-1<=n;j++)
{
rmqb[i][j]=max(rmqb[i-1][j],rmqb[i-1][j+(1<<(i-1))]);
rmqs[i][j]=min(rmqs[i-1][j],rmqs[i-1][j+(1<<(i-1))]);
}
}
int rmq_max(int l,int r)
{
int tmp=lg[r-l+1];
return max(rmqb[tmp][l],rmqb[tmp][r-(1<<tmp)+1]);
}
int rmq_min(int l,int r)
{
int tmp=lg[r-l+1];
return min(rmqs[tmp][l],rmqs[tmp][r-(1<<tmp)+1]);
}
void adde(int u,int v,int l)
{
edge[cnt].v=v;
edge[cnt].len=l;
edge[cnt].next=head[u];
head[u]=&edge[cnt++];
}
void dfs(int fa,int u)
{
int v;
node *p;
dp[u]=0;
for(p=head[u];p!=NULL;p=p->next)
{
v=p->v;
if(v==fa)
continue;
dfs(u,v);
p->md=dp[v]+p->len;
dp[u]=max(dp[u],dp[v]+p->len);
}
}
void getdis(int u,int v)
{
if(vis[v])
return ;
vis[v]=1;
node *p;
int md=0;
for(p=head[u];p!=NULL;p=p->next)
{
if(p->v!=v)
md=max(md,p->md);
}
for(p=head[v];p!=NULL;p=p->next)
if(p->v==u)
{
p->md=md+p->len;
dp[v]=max(dp[v],p->md);
}
for(p=head[v];p!=NULL;p=p->next)
getdis(v,p->v);
}
void predis()
{
node *p;
dfs(1,1);
vis[1]=1;
for(p=head[1];p!=NULL;p=p->next)
getdis(1,p->v);
}
int main()
{
int x,y,z,q,i,l,r,ans;
//freopen("in.txt","r",stdin);
pre_rmq();
while(scanf("%d%d",&n,&m),n||m)
{
init();
for(i=1;i<n;i++)
{
scanf("%d%d%d",&x,&y,&z);
adde(x,y,z);
adde(y,x,z);
}
predis();
rmq_init();
//for(i=1;i<=n;i++)
// printf("dis %d %d\n",i,dp[i]);
for(i=0;i<m;i++)
{
ans=1;
scanf("%d",&q);
l=r=1;
while(r<=n)
{
if(rmq_max(l,r)-rmq_min(l,r)<=q)
{
ans=max(ans,r-l+1);
r++;
}
else
l++;
}
printf("%d\n",ans);
}
}
return 0;
}