csu 1809 Parenthesis

题目见此

分析，把‘（’当成1， ‘）’当成-1， 计算前缀和sum。

记交换括号左边的序号为u， 右边为v，讨论左右括号：

1、s[u] == ‘(’ && s[v] == ‘)’ 那么[u, v - 1]的前缀和会全部-2

2、s[u] == ‘(’ && s[v] == ‘(’ 显然

3、s[u] == ‘)’ && s[v] == ‘(’ 那么[u, v - 1]的前缀和会全部+2

4、s[u] == ‘)’ && s[v] == ‘)’ 显然

对于一个括号序列，序列合法的要求是对∃sum[i]>=0 && sum[N]=0， 所以2，3，4的变换都不会改变序列的合法性。对于第一种情况，只要min(sum[i])>=2,i∈[u,v−1]即可。可以用ST或者线段树实现快速查询最小值。

/*****************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define   offcin        ios::sync_with_stdio(false)
#define   sigma_size    26
#define   lson          l,m,v<<1
#define   rson          m+1,r,v<<1|1
#define   slch          v<<1
#define   srch          v<<1|1
#define   sgetmid       int m = (l+r)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   pb            push_back
#define   fi            first
#define   se            second

const int    INF    = 0x3f3f3f3f;
const LL     INFF   = 1e18;
const double pi     = acos(-1.0);
const double inf    = 1e18;
const double eps    = 1e-9;
const LL     mod    = 1e9+7;
const int    maxmat = 10;
const ull    BASE   = 31;

/*****************************************************/

const int maxn = 2e5 + 5;
int N, Q;
char s[maxn];
int a[maxn];
int seg[maxn << 2], col[maxn << 2];
void PushUp(int v) {
    seg[v] = min(seg[slch], seg[srch]);
}
void build(int l, int r, int v) {
    if (l == r) seg[v] = a[l];
    else {
        sgetmid;
        build(lson);
        build(rson);
        PushUp(v);
    }
}
int query(int L, int R, int l, int r, int v) {
    if (L <= l && r <= R) return seg[v];
    sgetmid;
    int ans = INF;
    if (L <= m) ans = min(ans, query(L, R, lson));
    if (R >  m) ans = min(ans, query(L, R, rson));
    return ans;
}
int main(int argc, char const *argv[]) {
    int N, Q;
    while (~scanf("%d%d", &N, &Q)) {
        mem(seg, 0); mem(a, 0);
        scanf("%s", s);
        for (int i = 1; i <= N; i ++) {
            if (s[i - 1] == '(') a[i] = a[i - 1] + 1;
            else a[i] = a[i - 1] - 1;
        }
        build(1, N, 1);
        for (int i = 0; i < Q; i ++) {
            int u, v;
            scanf("%d%d", &u, &v);
            if (u > v) swap(u, v);
            if (s[u - 1] == '(' && s[v - 1] == ')') {
                int pos = query(u, v - 1, 1, N, 1);
                if (pos >= 2) puts("Yes");
                else puts("No");
            }
            else puts("Yes");
        }
    }
    return 0;
}