2011 Asia Beijing Regional Online Contest-1007 hdu4046 Panda

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题目有两个操作：

1. 查询一段区间内的wbw的个数

2. 修改某一个位置的字母

我们用树状数组记录sum[i] 表示以i之前出线的wbw的个数

当查询a，b区间时sum[b] - sum[a+1]就是a，b间wbw的个数

当更新某个点时，只需判定改变这个字母所带来的wbw个数的变化，并更新到树状数组中即可

#include <iostream>#include <cstdio>#include <string>#include <memory.h>using namespace std;const int M = 50005;char ch[M];int sum[M];int n,m;void INIT();void sol();void change(int k,int d);int SUM(int k);bool judge(int x);int lowbit(int k);int main(){    int t,cas = 1;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        scanf("%s",&ch);        INIT();        printf("Case %d:\n",cas++);        sol();    }}void INIT(){    memset(sum,0,sizeof(sum));    for(int i = 2;i < n;i++)    {        if(ch[i-2] == 'w' && ch[i-1] == 'b' && ch[i] == 'w')            change(i,1);    }    return;}void sol(){    int a,b,c;    char d;    for(int i = 0;i < m;i++)    {        scanf("%d%",&a);        if(a == 0)        {            scanf("%d%d",&b,&c);            if(c - b <= 1)                printf("0\n");            else                printf("%d\n",SUM(c)- SUM(b + 1));        }        else if(a == 1)        {            scanf("%d",&b);            cin >> d;            if(d != ch[b])            {                if(judge(b))                    change(b,-1);                if(judge(b+1))                    change(b+1,-1);                if(judge(b+2))                    change(b+2,-1);                ch[b] = d;                if(judge(b))                    change(b,1);                if(judge(b+1))                    change(b+1,1);                if(judge(b+2))                    change(b+2,1);            }        }    }}bool judge(int x){    if(x < 2)return false;    if(ch[x-2] == 'w' && ch[x-1] == 'b' && ch[x] == 'w')        return true;    else return false;}int lowbit(int k){return k&(-k);}void change(int k,int d){    while(k < M)    {        sum[k] += d;        k += lowbit(k);    }    return;}int SUM(int k){    int s = 0;    while(k > 0)    {        s += sum[k];        k-= lowbit(k);    }    return s;}