hdu 4123 Bob’s Race (树的直径+尺取+rmq)(珍爱生命,远离log)

Posted by 111qqz on Sunday, July 17, 2016

TOC

hdu 4123 题目链接

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。。。

2016-07-17 23-19-46 的屏幕截图
2016-07-17 23-26-58 的屏幕截图

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;
}
v

「真诚赞赏,手留余香」

111qqz的小窝

真诚赞赏,手留余香

使用微信扫描二维码完成支付