hdu 4123 Bob’s Race (树的直径+尺取+rmq)(珍爱生命，远离log)
题意：一棵树，定义d[i]为点i到树上某点的最大距离。。。给出若干查询，每个查询一个x,问最多能有多少点满足这些点中，最大的d与最小的d的差小于等于x.要求这些点的编号必须是连续的。
思路：可以三遍bfs处理出所有点的d...
由于不能排序。。。所以就是尺取+rmq....
然而神Tm TLE.....
这复杂度还TLe...
结果最后发现是。。。log运算的常数太大被卡。。。
所以做法是先预处理一下。。。嗯。。。。
珍爱生命，远离log!
珍爱生命，远离log!
珍爱生命，远离log!
/* ***********************************************
Author :111qqz
Created Time :2016年07月17日 星期日 19时37分55秒
File Name :code/hdu/4123.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <string>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
#define log2 0.693147
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E5+7; //双向边。。。
int n,Q;
int head[N];
int d[5][N];
bool vis[N];
int ans;
int cnt;
int dp[N][30],dp2[N][30];
int beg,lst;
int LOG[N+10];
struct Edge
{
int v;
int w;
int nxt;
}edge[N];
void addedge( int u,int v,int w)
{
edge[cnt].v = v;
edge[cnt].w = w;
edge[cnt].nxt = head[u];
head[u] = cnt;
cnt++;
}
void rmq_init()
{
for ( int i = 1 ; i <= n ; i++)
dp[i][0] = dp2[i][0] = d[0][i];
for ( int j = 1 ; (1<<j)<=n; j++)
for ( int i = 1 ; i + (1<<j) -1 <= n ; i++)
{
dp[i][j] = max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
dp2[i][j] = min(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
}
}
int rmq( int l,int r)
{
if (l>r) return 0;
int k = log(double (r-l+1)/log2);
//int k = LOG[r-l+1];
//int k = 0;
//while (1<<(k+1)<=r-l+1) k++;
int mx = max(dp[l][k],dp[r-(1<<k)+1][k]);
int mn = min(dp2[l][k],dp2[r-(1<<k)+1][k]);
return mx - mn;
}
void init()
{
ms(head,-1);
cnt = 0;
}
void bfs( int s,int k)
{
ms(d[k],0x3f);
ms(vis,false);
queue<int>q;
q.push(s);
vis[s] = true;
d[0][s] = 0;
while (!q.empty())
{
int u = q.front();
q.pop();
for ( int i = head[u] ; i !=-1 ; i = edge[i].nxt)
{
int v = edge[i].v;
int w = edge[i].w;
if (vis[v]) continue;
vis[v] = true;
d[k][v] = d[k][u] + w;
q.push(v);
}
}
}
void ruler(int k)
{
int head = 1;
int tail = 1;
ans = 0 ;
// for ( int i = 1 ; i <= n ; i++) cout<<"d[0][i]:"<<d[0][i]<<endl;
// cout<<"k:"<<k<<endl;
while (tail<=n)
{
int cur = rmq(head,tail);
// cout<<"head:"<<head<<" tail:"<<tail<<" cur:"<<cur<<endl;
while (head<tail&&rmq(head,tail)>k) head++;
cur = rmq(head,tail);
if (cur<=k) ans = max(ans,tail-head);
tail++;
}
ans++;
}
void ruler2 ( int k)
{
ans = 0 ;
int id = 1;
for ( int i = 1 ;i <= n ; i++)
{
while (id<=i&&rmq(id,i)>k) id++;
ans =max(ans,i-id+1);
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("code/in.txt","r",stdin);
#endif
LOG[0] = -1;
for ( int i = 1 ; i < 2* N ; i++) LOG[i] =LOG[i>>1]+1;
while (scanf("%d%d",&n,&Q),n||Q)
{
// cout<<log(2.0)<<endl;
init();
for ( int i = 1 ; i <= n-1 ; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
bfs(1,0);
int far = 0 ;
for ( int i = 1 ; i <= n ; i++)
if (d[0][i]>far)
{
far = d[0][i];
lst = i ;
}
far = 0 ;
bfs(lst,1);
for ( int i = 1 ;i <= n ; i++)
if (d[1][i]>far)
{
far = d[1][i];
beg = i ;
}
bfs(beg,2);
for ( int i = 1 ; i <= n ; i++)
d[0][i] = max(d[1][i],d[2][i]);
rmq_init();
while (Q--)
{
int x;
scanf("%d",&x);
ruler(x);
printf("%d\n",ans);
}
}
#ifndef ONLINE_JUDGE
fclose(stdin);
#endif
return 0;
}
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