hdu 4046 Panda

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

　　简单线段树【单点修改，查询区间】，合并两个目标区间，查找区间中目标子串的个数，子串可以有重叠的部分！

AC代码：

#include <cstring>
#include <cstdio>
#include <cassert>
#include <algorithm>

using namespace std;

const int maxn = 50001;
int cnt[maxn << 2];

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

char str[maxn];

void up(int rt, int l, int m, int r) {
    cnt[rt] = cnt[rt << 1] + cnt[rt << 1 | 1];
    if (m - l) {
        if (str[m - 1] == 'w' && str[m] == 'b' && str[m + 1] == 'w') {
            cnt[rt]++;
        }
        if (r - m > 1) {
            if (str[m] == 'w' && str[m + 1] == 'b' && str[m + 2] == 'w') {
                cnt[rt]++;
            }
        }
    }
}

void build(int l, int r, int rt) {
    if (l == r) {
        cnt[rt] = 0;

        return ;
    }
    int m = (l + r) >> 1;

    build(lson);
    build(rson);
    up(rt, l, m, r);
}

void update(int pos, char ch, int l, int r, int rt) {
    if (l == r) {
        str[pos] = ch;

        return ;
    }
    int m = (l + r) >> 1;

    if (pos <= m) update(pos, ch, lson);
    else update(pos, ch, rson);
    up(rt, l, m, r);
}

struct Answer {
    int l;
    int r;
    int cnt;
};

Answer query(int L, int R, int l, int r, int rt) {
    Answer ret;

    if (L <= l && r <= R) {
        ret.l = l;
        ret.r = r;
        ret.cnt = cnt[rt];

        return ret;
    }
    int m = (l + r) >> 1;

    if (L <= m) {
        ret = query(L, R, lson);
        if (m < R) {
            Answer tmp = query(L, R, rson);
            ret.cnt += tmp.cnt;

            int t = ret.r;

            if (ret.r - ret.l) {

                if (str[t - 1] == 'w' && str[t] == 'b' && str[t + 1] == 'w') {
                    ret.cnt++;
                }
                if (tmp.r - tmp.l) {
                    if (str[t] == 'w' && str[t + 1] == 'b' && str[t + 2] == 'w') {
                        ret.cnt++;
                    }
                }
            } else {
                if (tmp.r - tmp.l) {
                    if (str[t] == 'w' && str[t + 1] == 'b' && str[t + 2] == 'w') {
                        ret.cnt++;
                    }
                }
            }
            ret.r = tmp.r;
        }
    } else if (m < R) {
        ret = query(L, R, rson);
    }

    return ret;
}

void deal() {
    int n,m;

    scanf("%d%d", &n, &m);
    scanf("%s", str);

    build(0, n - 1, 1);
    while (m--) {
        int op, l, r;
        char ch[3];

        scanf("%d", &op);
        if (op) {
            scanf("%d %s", &l, ch);
            update(l, ch[0], 0, n - 1, 1);
        } else {
            scanf("%d%d", &l, &r);
            printf("%d\n", query(l, r, 0, n - 1, 1).cnt);
        }
    }
}

int main() {
    int T;

    scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
        printf("Case %d:\n", i);
        deal();
    }

    return 0;
}

——written by Lyon