Bob’s Race
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1994 Accepted Submission(s): 619
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 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
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
3
3
3
5
Source
首先是两遍dfs,预处理出每个结点到叶子结点的巨大距离。
然后使用rmq来查询区间的最大最小值。
每次查询扫描一遍就可以了、
/* ***********************************************
Author :kuangbin
Created Time :2013-11-8 16:56:11
File Name :E:\2013ACM\专题强化训练\区域赛\2011福州\C.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = ;
struct Edge
{
int to,next;
int w;
}edge[MAXN*];
int head[MAXN],tot;
void init()
{
tot = ;
memset(head,-,sizeof(head));
}
void addedge(int u,int v,int w)
{
edge[tot].to = v;
edge[tot].w = w;
edge[tot].next = head[u];
head[u] = tot++;
}
int maxn[MAXN],smaxn[MAXN];
int maxid[MAXN],smaxid[MAXN];
void dfs1(int u,int pre)
{
maxn[u] = smaxn[u] = maxid[u] = smaxid[u] = ;
for(int i = head[u];i != -;i = edge[i].next)
{
int v = edge[i].to;
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 != -;i = edge[i].next)
{
int v = edge[i].to;
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[MAXN]; int dp1[MAXN][];
int dp2[MAXN][];
int mm[MAXN];
void initRMQ(int n)
{
mm[] = -;
for(int i = ;i <= n;i++)
{
mm[i] = ((i&(i-)) == )?mm[i-]+:mm[i-];
dp1[i][] = a[i];
dp2[i][] = a[i];
}
for(int j = ;j <= mm[n];j++)
for(int i = ;i + (<<j) - <= n;i++)
{
dp1[i][j] = max(dp1[i][j-],dp1[i + (<<(j-))][j-]);
dp2[i][j] = min(dp2[i][j-],dp2[i + (<<(j-))][j-]);
}
}
int rmq(int x,int y)
{
int k = mm[y-x+];
return max(dp1[x][k],dp1[y-(<<k)+][k]) - min(dp2[x][k],dp2[y-(<<k)+][k]);
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n,m;
int u,v,w;
int Q;
while(scanf("%d%d",&n,&m) == )
{
if(n == && m == )break;
init();
for(int i = ;i < n;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
dfs1(,);
dfs2(,);
for(int i = ;i <= n;i++)
a[i] = maxn[i];
initRMQ(n);
while(m--)
{
scanf("%d",&Q);
int ans = ;
int id = ;
for(int i = ;i <= n;i++)
{
while(id <= i && rmq(id,i) > Q)id++;
ans = max(ans,i-id+);
}
printf("%d\n",ans);
}
}
return ;
}