ZR#331. 【18 提高 3】括号序列(栈)

题意

挺神仙的。首先$60$分暴力是比较好打的。

就是枚举左端点，看右端点能否是$0$

但是这样肯定是过不了的，假如我们只枚举一次，把得到的栈记录下来

那么若区间$(l, r)$是可行的，那么$s_{l - } = s_r$，证明自己yy一下吧。。

然后就是字符串hash乱搞了。。

#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
#include<map>
#define LL long long
#define ull unsigned long long
using namespace std;
const int MAXN = 1e6 + , mod = 1e9 + ;
inline LL read() {
    char c = getchar(); LL x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
char a[MAXN], s[MAXN];
int top = ;
ull base = , base2 = , po1[MAXN], po2[MAXN];
map<ull, LL> mp;
int main() {
    scanf("%s", a + );
    LL N = strlen(a + ), ans = ;
    ull sum = , sum2 = ;
    mp[] = ; po1[] = ; po2[] = ;
    for(int i = ; i <= N; i++) po1[i] = po1[i - ] * base, po2[i] = po2[i - ] * base2;
    for(int i = ; i <= N; i++) {
        if(top && a[i] == s[top]) {
            top--;
            sum -= po1[top] * (a[i] - 'a' + );
            sum2 -= po2[top] * (a[i] - 'a' + );

        }
        else {
            sum += po1[top] * (a[i] - 'a' + );
            sum2 += po2[top] * (a[i] - 'a' + ) ;
            s[++top] = a[i];
        }
        ans += mp[(sum << ) + sum2];
        mp[(sum << ) + sum2]++;
     //   printf("%d\n", ans);
    }
    printf("%lld", ans);
    return ;
}
/*
abaababababbbbbaavbaaaabbbaaaabbabbbaabbabbb
*/