hdu 1199 Color the Ball

Color the Ball

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3081    Accepted Submission(s): 762

Problem Description

There are infinite balls in a line (numbered 1 2 3 ....), and initially all of them are paint black. Now Jim use a brush paint the balls, every time give two integers a b and follow by a char 'w' or 'b', 'w' denotes the ball from a to b are painted white, 'b' denotes that be painted black. You are ask to find the longest white ball sequence.

 

Input

First line is an integer N (<=2000), the times Jim paint, next N line contain a b c, c can be 'w' and 'b'.

There are multiple cases, process to the end of file.

 

Output

Two integers the left end of the longest white ball sequence and the right end of longest white ball sequence (If more than one output the small number one). All the input are less than 2^31-1. If no such sequence exists, output "Oh, my god".

 

Sample Input

3

1 4 w

8 11 w

3 5 b

 

Sample Output

8

11

 

Author

ZHOU, Kai

 

Source

ZOJ Monthly, February 2005

 

Recommend

Ignatius.L

开始求错了东西、好惨，Wa的惨不忍睹

离散化
线段树更新，最后更新到底 NlogN
然后对树底暴力进行统计 （N）
细节：
把每个点拆成2份
奇数代表区间 偶数代表点

#include <iostream>
#include <map>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 4003
int st[N<<3];
struct record
{
    int l,r;
    char c;
}rc[N>>1];
int id[N<<1],ls[N],cnt;
int bf(int n)
{
    int l=1,r=cnt-1,m;
    while(l<=r)
    {
        m=(l+r)>>1;
        if(ls[m]>n) r=m-1;
        else if(ls[m]<n) l=m+1;
        else return m;
    }
    return 0;
}
void build(int l,int r,int k)
{
    st[k]=0;
    if(l==r)
    {
        id[cnt++]=k;
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void down(int &k)
{
    st[k<<1]=st[k];
    st[k<<1|1]=st[k];
    st[k]=-1;
}
int L,R,flag;
void  update(int l,int r,int k)
{
    if(L<=l&&R>=r)
    {
       st[k]=flag;
       return ;
    }
    if(st[k]!=-1)
     down(k);
    int m=(l+r)>>1;
    if(L<=m) update(lson);
    if(R>m)  update(rson);
}
void over(int l,int r,int k)
{
    if(l==r) return;
    if(st[k]!=-1)
     down(k);
    int m=(l+r)>>1;
     over(lson);
     over(rson);
}
int main()
{
    int n;
    int i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        for(k=1,i=0;i<n;i++)
        {
            scanf("%d %d %c",&rc[i].l,&rc[i].r,&rc[i].c);
            ls[k++]=rc[i].l;
            ls[k++]=rc[i].r;
        }
        sort(ls+1,ls+k);
        n=(n<<1);
        for(k=1,i=2;i<=n;i++)
         if(ls[i]!=ls[k])
          ls[++k]=ls[i];
        cnt=1;
        build(1,k<<1,1);
        n=(n>>1);
        for(i=0;i<n;i++)
        {
            L=bf(rc[i].l)<<1;
            R=bf(rc[i].r)<<1;
            flag=rc[i].c=='w'?1:0;
            update(1,k<<1,1);
        }
        over(1,k<<1,1);
      //  for(i=1;i<cnt;i++)
       //    printf("%d %d ->",i,st[id[i]]);
        int rcl=-1,rcr=-1,rclen=-1;
        for(i=1;i<cnt;i++)
         if(st[id[i]])
         {
             int l,r;
              if(i%2)
                 l=ls[i/2]+1;开始写成了-1，白白贡献了个 WA
              else
                l=ls[i/2];
            //printf("%d %d\n",i,l);
            for(j=i+1;j<cnt;j++)
                if(st[id[j]]==0)
                {
                    if(j%2)
                        if(st[id[j+1]]&&ls[j/2+1]-ls[j/2]==1)
                          continue;
                         else
                         {
                             r=ls[j/2];
                             break;
                         }
                    else
                        {
                            r=ls[j/2]-1;
                            break;
                        }
                }
              if(j==cnt) r=ls[j/2];
              if(r-l>rclen)
              {
                  rclen=r-l;
                  rcl=l;
                  rcr=r;
              }
           i=j;
         }

         if(rcl==-1)
          printf("Oh,my god\n");
        else
          printf("%d %d\n",rcl,rcr);
    }
    return 0;
}