题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4046
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?
题意描述：给出一个仅包含b和w字符的字符串，然后很多操作：
type=0 : l r，求出区间[l,r]包含wbw这样字符三元组的个数；
type=1 : k ch，把第k个字符改为ch。
算法分析：树状数组。hdu上听说这道题数据很水，暴力就可以过，那么这道题就没有情趣了。下面简单介绍一下树状数组的做法：
首先遍历字符串，如果在 i 位置发现[i,i+2]连续字符串为wbw，则运用树状数组的方法add(i,1)，修改字符的时候，判断修改前后wbw的数量就可以了。
然后询问这里有点绕，如果询问[l,r]的话，把r的位置往前移动两个位置，l往前移动一个位置。比如字符串为wbwbw(第一个位置序号为1)，询问[1,3]的个数，我们就相当于询问[0,1]，这样避免了第二个wbw影响了这个询问。还有，需要特判一下改后的这个区间是否l<r即可。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #define inf 0x7fffffff
8 using namespace std;
9 typedef long long LL;
10 const int maxn=50000+10;
11
12 int n,m;
13 int c[maxn];
14 char str[maxn];
15
16 int lowbit(int u) {return u&(-u); }
17 void add(int i,int dd)
18 {
19 while (i<=maxn)
20 {
21 c[i] += dd;
22 i += lowbit(i);
23 }
24 }
25
26 int sum(int i)
27 {
28 int ret=0;
29 while (i>0)
30 {
31 ret += c[i];
32 i -= lowbit(i);
33 }
34 return ret;
35 }
36
37 int main()
38 {
39 int t,ncase=1;scanf("%d",&t);
40 while (t--)
41 {
42 scanf("%d%d",&n,&m);
43 scanf("%s",str+1);
44 printf("Case %d:\n",ncase++);
45 memset(c,0,sizeof(c));
46 int len=strlen(str+1);
47 for (int i=1 ;i<=len-2 ;i++)
48 {
49 if (str[i]==‘w‘ && str[i+1]==‘b‘ && str[i+2]==‘w‘)
50 add(i,1);
51 }
52 char ch[3];
53 int l,r;
54 int type;
55 for (int i=0 ;i<m ;i++)
56 {
57 scanf("%d",&type);
58 if (type==0)
59 {
60 scanf("%d%d",&l,&r);
61 if (l>=r-1) printf("0\n");
62 else printf("%d\n",sum(r-1)-sum(l));
63 }
64 else
65 {
66 scanf("%d%s",&l,ch);
67 if (l+1-2>=1 && l+1<=n)
68 {
69 if (str[l+1-2]==‘w‘&&str[l+1-1]==‘b‘)
70 {
71 if (str[l+1]==‘w‘ && ch[0]==‘b‘)
72 add(l+1-2,-1);
73 else if (str[l+1]==‘b‘&&ch[0]==‘w‘)
74 add(l+1-2,1);
75 }
76 }
77 if (l+1-1>=1 && l+1+1<=n)
78 {
79 if (str[l+1-1]==‘w‘&&str[l+1+1]==‘w‘)
80 {
81 if (str[l+1]==‘b‘&&ch[0]==‘w‘)
82 add(l+1-1,-1);
83 else if (str[l+1]==‘w‘&&ch[0]==‘b‘)
84 add(l+1-1,1);
85 }
86 }
87 if (l+1>=1 && l+1+2<=n)
88 {
89 if (str[l+1+1]==‘b‘&&str[l+1+2]==‘w‘)
90 {
91 if (str[l+1]==‘w‘&&ch[0]==‘b‘)
92 add(l+1,-1);
93 else if (str[l+1]==‘b‘&&ch[0]==‘w‘)
94 add(l+1,1);
95 }
96 }
97 str[l+1]=ch[0];
98 }
99 }
100 }
101 return 0;
102 }