hdu4046(线段树)

http://acm.hdu.edu.cn/showproblem.php?pid=4046

Panda

Time Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1516 Accepted Submission(s): 511

Problem Description

When I wrote down this letter, you may have been on the airplane to U.S.
We have known for 15 years, which has exceeded one-fifth of my whole life. I still remember the first time we went to the movies, the first time we went for a walk together. I still remember the smiling face you wore when you were dressing in front of the mirror. I love your smile and your shining eyes. When you are with me, every second is wonderful.
The more expectation I had, the more disappointment I got. You said you would like to go to U.S.I know what you really meant. I respect your decision. Gravitation is not responsible for people falling in love. I will always be your best friend. I know the way is difficult. Every minute thinking of giving up, thinking of the reason why you have held on for so long, just keep going on. Whenever you’re having a bad day, remember this: I LOVE YOU.
I will keep waiting, until you come back. Look into my eyes and you will see what you mean to me.
There are two most fortunate stories in my life: one is finally the time I love you exhausted. the other is that long time ago on a particular day I met you.
Saerdna.

It comes back to several years ago. I still remember your immature face.
The yellowed picture under the table might evoke the countless memory. The boy will keep the last appointment with the girl, miss the heavy rain in those years, miss the love in those years. Having tried to conquer the world, only to find that in the end, you are the world. I want to tell you I didn’t forget. Starry night, I will hold you tightly.

Saerdna loves Panda so much, and also you know that Panda has two colors, black and white.
Saerdna wants to share his love with Panda, so he writes a love letter by just black and white.
The love letter is too long and Panda has not that much time to see the whole letter.
But it's easy to read the letter, because Saerdna hides his love in the letter by using the three continuous key words that are white, black and white.
But Panda doesn't know how many Saerdna's love there are in the letter.
Can you help Panda?

 

Input

An integer T means the number of test cases T<=100
For each test case:
First line is two integers n, m
n means the length of the letter, m means the query of the Panda. n<=50000,m<=10000
The next line has n characters 'b' or 'w', 'b' means black, 'w' means white.
The next m lines
Each line has two type
Type 0: answer how many love between L and R. (0<=L<=R
Type 1: change the kth character to ch(0<=k

 

Output

For each test case, output the case number first.
The answer of the question.

 

Sample Input

2 5 2 bwbwb 0 0 4 0 1 3 5 5 wbwbw 0 0 4 0 0 2 0 2 4 1 2 b 0 0 4

 

Sample Output

Case 1: 1 1 Case 2: 2 1 1 0

 

Source

The 36th ACM/ICPC Asia Regional Beijing Site —— Online Contest

 

Recommend

lcy

果断参考大神的代码。。 看了半天终于看懂了。

参考出处： http://hi.baidu.com/517145/item/f3d3c539f3d8409ef4e4ad0e

作者：未知。

题意：给你一个字符串，由'w' 和 'b' 组成，对该字符串有俩个操作，
当输入为0时，询问该区间[a,b] 内有多少个串 为 "wbw";
当输入为0时，将下标为k的字符改为输入的字符；
注意：这里的a，b，k表示的都是字符串的下标，也就是取值范围为[0,n-1];
分析：
对这种区间修改或询问的题目，首先想到的就是线段树或者树状数组了，不过还是线段树熟悉一点，
我用线段树做的 ，关键还是构造模型，区间内每一个点的值表示的是什么？
我是这样表示的，每一个点表示的是以该下标结束的长度为3的子串是否为"wbw",
若是，则等于1，否则等于0，保存在数组a[]中。
以，当进行区间询问时，对[a,b]的询问则应该改为对区间[a+2,b]，当然这里对a+2，和b的大小还是要讨论的；
当进行点修改时，则可能要修改三个点，因为改动一个字符时，
会影响到三个长度为3的子串，所以有可能进行三次修改（之前在这里对条件的判断似乎不太严格，一直WA）；

#include<stdio.h>
#include<iostream>
#include<memory.h>
#include<string.h>
using namespace std ;
#define MAX 50012
struct node
{
    int l ,r ,c ;
};
struct node tree [MAX * 4 ];
#define Mid(idx) (tree[idx].l+tree[idx].r)>>1
int num [MAX * 5 ];
void buildTree ( int idx , int l , int r )
{
      tree [idx ].l =l ;
      tree [idx ].r =r ;
      tree [idx ].c =- 1 ;
    if (r ==l )
    {
              tree [idx ].c =num [l ];
        return ;
    }
    int mid =(l +r )>> 1 ;
      buildTree (idx << 1 ,l ,mid );
      buildTree (idx << 1 | 1 ,mid + 1 ,r );
      tree [idx ].c =tree [idx << 1 ].c +tree [idx << 1 | 1 ].c ;
}
void Insert ( int idx , int i , int c )
{
    if (tree [idx ].l ==i &&i ==tree [idx ].r )
    {
              tree [idx ].c =c ;
        return ;
    }
    int mid =Mid (idx );
    if (i >mid ) Insert (idx << 1 | 1 ,i ,c ); //i>mid 就可以了，我土血啊，
    else Insert (idx << 1 ,i ,c );
      tree [idx ].c =tree [idx << 1 ].c +tree [idx << 1 | 1 ].c ; //计算区间和
}
int querySum ( int idx , int l , int r )
{
    if (tree [idx ].l ==l &&tree [idx ].r ==r )
    {
        return tree [idx ].c ;
    }
    int mid =Mid (idx );
    if (mid >=r ) return querySum (idx << 1 ,l ,r );
    else if (l >mid ) return querySum (idx << 1 | 1 ,l ,r );
    else
    {
        return querySum (idx << 1 ,l ,mid )+querySum (idx << 1 | 1 ,mid + 1 ,r );
    }
}
int main ()
{
    int T ,n ,m ,i ,type ,l ,r ,indx ,iCase = 0 ;
    char str [MAX * 5 ],temp ;
      scanf ( "%d" ,&T );
    while (T --)
    {
              memset (num , 0 , sizeof (num ));

              scanf ( "%d%d" ,&n ,&m );
              scanf ( "%s" ,&str );
              num [ 1 ]= 0 ;
              printf ( "Case %d:\n" ,++iCase );
        for (i = 2 ; i <n ; i ++)
        {
            if (str [i - 2 ]== 'w' &&str [i - 1 ]== 'b' &&str [i ]== 'w' )
                              num [i ]= 1 ;
            else 
                              num [i ]= 0 ;
        }
              num [n ]= 0 ;
              buildTree ( 1 , 1 ,n );
        for (i = 0 ; i <m ; i ++)
        {
                      scanf ( "%d" ,&type );
            if ( 0 ==type )
            {
                              scanf ( "%d%d" ,&l ,&r );
                if (l + 2 >r ) //询问的区间长度<3直接输出0
                {
                                      cout << 0 <<endl ;
                    continue ;
                } //否则从l+2开始询问
                              cout <<querySum ( 1 ,l + 2 ,r )<<endl ;
            }
            else
            {
                              scanf ( "%d %c" ,&indx ,&temp );
                if (temp ==str [indx ]) continue ; //若要就该的字符跟原先相同，则不用修改了
                if (indx >= 2 &&str [indx - 2 ]== 'w' &&str [indx - 1 ]== 'b' &&str [indx ]== 'w' )
                {
                                      Insert ( 1 ,indx , 0 );
                                      num [indx ]= 0 ;
                }
                if (indx >= 2 &&str [indx - 2 ]== 'w' &&str [indx - 1 ]== 'b' &&temp == 'w' )
                {
                                              Insert ( 1 ,indx , 1 );
                                              num [indx ]= 1 ;
                }
                if (indx >= 1 &&indx + 1 <n &&str [indx - 1 ]== 'w' &&str [indx ]== 'b' &&str [indx + 1 ]== 'w' )
                {
                                      Insert ( 1 ,indx + 1 , 0 );
                                      num [indx + 1 ]= 0 ;
                }
                if (indx >= 1 &&indx + 1 <n &&str [indx - 1 ]== 'w' &&temp == 'b' &&str [indx + 1 ]== 'w' )
                {
                                              Insert ( 1 ,indx + 1 , 1 );
                                              num [indx + 1 ]= 1 ;
                }
                if (indx >= 0 &&indx + 2 <n &&str [indx ]== 'w' &&str [indx + 1 ]== 'b' &&str [indx + 2 ]== 'w' )
                {
                                      Insert ( 1 ,indx + 2 , 0 );
                                      num [indx + 2 ]= 0 ;
                }
                if (indx >= 0 &&indx + 2 <n &&temp == 'w' &&str [indx + 1 ]== 'b' &&str [indx + 2 ]== 'w' )
                {
                                              Insert ( 1 ,indx + 2 , 1 );
                                              num [indx + 2 ]= 1 ;
                }
                              str [indx ]=temp ;
            }
        }
    }
    return 0 ;
}