线段树 HDU 4046 panda

来源：互联网发布：linux安装ha 编辑：程序博客网时间：2024/04/29 11:29

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

代码风格：www.notonlysuccess.com

题目大意：求wbw出现的个数

算法：线段树，区间合并

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define lson l, m, rt << 1#define rson m+1, r, rt << 1 | 1#define mid int m = (r+l) >> 1int cnt[454545];char s[465456];void PushUp(int L, int R, int m, int rt){    cnt[rt] = cnt[rt << 1] + cnt[rt << 1 | 1];    if(L <= m && m+2 <= R)        if(s[m] == 'w' && s[m+1] == 'b' && s[m+2] == 'w')            cnt[rt] ++;    if(L <= m-1 && m + 1 <= R)    {        if(s[m-1] == 'w' && s[m] == 'b' && s[m+1] == 'w')        {            cnt[rt] ++;        }    }}void build(int l, int r, int rt){    if(r == l)        return ;    mid ;    build(lson);    build(rson);    PushUp(l, r, m, rt);}void update(int d, int l, int r, int rt){    if(l == r)    {        return ;    }    mid ;    if(d <= m)        update(d, lson);    else update(d, rson);    PushUp(l, r, m, rt);}int minz(int a, int b){    return a < b ? a : b;}int maxz(int a, int b){    return a > b ? a : b;}int query(int L, int R, int l, int r, int rt){    if(L <= l && r <= R)        return cnt[rt];    mid ;    int ret  = 0;    if(R <= m)        return query(L, R, lson);    else if(L > m)        return query(L, R, rson);    else        ret += query(L, R, lson) + query(L, R, rson);    if(maxz(l, L) <= m && m+2 <= minz(r, R))        if(s[m] == 'w' && s[m+1] == 'b' && s[m+2] == 'w')            ret ++;    if(maxz(l, L) <= m-1 && m + 1 <= minz(r, R))        if(s[m-1] == 'w' && s[m] == 'b' && s[m+1] == 'w')            ret ++;    return ret ;}int main(){    int n, m, T;    int a, b, c;    int ica = 1;    char op[45];    scanf("%d", &T);    while(T --){        printf("Case %d:\n", ica ++);        memset(s, 0, sizeof(s));        memset(cnt, 0, sizeof(cnt));        scanf("%d%d", &n, &m);        scanf("%s", s);        build(0, n-1, 1);        while(m --){            scanf("%d%d", &a, &b);            if(a == 1)            {                scanf("%s", op);                s[b] = op[0];                update(b, 0, n-1, 1);            }            else            {                scanf("%d", &c);                printf("%d\n", query(b, c, 0, n-1, 1));            }        }    }    return 0;}