Parenthesis
Problem Description：
Bobo has a balanced parenthesis sequence P=p1 p2…pn of length n and q questions.
The i-th question is whether P remains balanced after pai and pbi swapped. Note that questions are individual so that they have no affect on others.
Parenthesis sequence S is balanced if and only if:
-
S is empty;
-
or there exists balanced parenthesis sequence A,B such that S=AB;
-
or there exists balanced parenthesis sequence S' such that S=(S').
Input：
The input contains at most 30 sets. For each set:
The first line contains two integers n,q (2≤n≤105,1≤q≤105).
The second line contains n characters p1 p2…pn.
The i-th of the last q lines contains 2 integers ai,bi (1≤ai,bi≤n,ai≠bi).
Output：
For each question, output "Yes" if P remains balanced, or "No" otherwise.
Sample Input：
4 2
(())
1 3
2 3
2 1
()
1 2
Sample Output：
No
Yes
No
【题目链接】CSU 1809 Parenthesis
【题目类型】线段树+前缀和
&题意：
给了一个平衡的括号序列s（平衡是指括号匹配正常）现在q次询问，每次输入两个数a、b;问将s[a]和s[b]交换后是否仍然平衡（即是否正常匹配）平衡则输出“Yes”，否则输出“No”
&题解：
这种括号的题目有一种经典的套路就是遇到'('加一，遇到')'减一，这样整个序列最后的前缀和一定是非负的，同样的这里贪心一下就会发现只有把'(' 和')'交换的时候才会出问题，这时我们就要想一下了，有了括号的套路，你只给一个数组赋值正负一是没什么意义的，所以你要求他的前缀和，放在pre数组中，这样假设每次修改的区间是u，v，也就是只有当s[u] == '(' && s[v] == ')'才需要判断，且u<=v，那么要判断什么呢？你想，正确情况下他的pre是不会有<0的情况的，但如果换了之后，前面u的位置换成了'('就要减1，后面v的位置是加1的，所以不用管，这时候pre在区间[u,v-1]的最小值>=2才可以。
【时间复杂度】O(nlogn)
&代码：
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
#define SII(N,M) scanf("%d %d",&(N),&(M))
#define rep(i,b) for(int i=0;i<(b);i++)
#define rez(i,a,b) for(int i=(a);i<=(b);i++)
const int MAXN = 100000 + 5 ;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int sum[MAXN << 2], pre[MAXN];
inline void Pushmin(int rt) {
sum[rt] = min(sum[rt << 1], sum[rt << 1 | 1]);
}
void Build(int l, int r, int rt) {
if (l == r) {
sum[rt] = pre[l];
return;
}
int m = (l + r) >> 1;
Build(lson);
Build(rson);
Pushmin(rt);
}
int Query(int L, int R, int l, int r, int rt) {
if (L <= l && r <= R) {
return sum[rt];
}
int m = (l + r) >> 1;
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 n, q;
char s[MAXN];
void Solve() {
while (~SII(n, q)) {
scanf("%s", s + 1);
rez(i, 1, n) if (s[i] == '(') pre[i] = pre[i - 1] + 1;
else pre[i] = pre[i - 1] - 1;
Build(1, n, 1);
rep(i, q) {
int u, v;
SII(u, v);
if (v < u) swap(u, v);
if (s[u] == '(' && s[v] == ')') {
int d = Query(u, v - 1, 1, n, 1);
if (d < 2) puts("No");
else puts("Yes");
}
else puts("Yes");
}
}
}
int main() {
Solve();
return 0;
}