Time Limit: 5 Sec Memory Limit: 128 Mb Submitted: 2291 Solved: 622
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:
1. S is empty;
2. or there exists balanced parenthesis sequence A,B such that S=AB;
3. 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
Hint
Source
湖南省第十二届大学生计算机程序设计竞赛
//题意: n 长字符串，m次询问，一开始括号是匹配的，问 a ，b 位置的字符调换后，是否匹配
//题解:网上很多的暴力超时代码没看懂，这题，如果，( 看成 1 ，) 看成 -1 ，其实就是求 a -- (b-1) 中前缀和最小的，在调换后是否能满足要求即可，写了个线段树求区间最小
1 # include <cstdio> 2 # include <cstring> 3 # include <cstdlib> 4 # include <iostream> 5 # include <vector> 6 # include <queue> 7 # include <stack> 8 # include <map> 9 # include <bitset> 10 # include <sstream> 11 # include <set> 12 # include <cmath> 13 # include <algorithm> 14 #pragma comment(linker,"/STACK:102400000,102400000") 15 using namespace std; 16 #define LL long long 17 #define lowbit(x) ((x)&(-x)) 18 #define PI acos(-1.0) 19 #define INF 0x3f3f3f3f3f3f3f3f 20 #define eps 1e-8 21 #define MOD 1000000007 22 23 inline int scan() { 24 int x=0,f=1; char ch=getchar(); 25 while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();} 26 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} 27 return x*f; 28 } 29 inline void Out(int a) { 30 if(a<0) {putchar('-'); a=-a;} 31 if(a>=10) Out(a/10); 32 putchar(a%10+'0'); 33 } 34 #define MX 100005 35 /**************************/ 36 struct Tree 37 { 38 int l,r; 39 int m; 40 }tree[MX*4]; 41 42 int st[MX]; 43 char str[MX]; 44 45 void build_tree(int l,int r,int k) 46 { 47 tree[k] = (Tree){l,r,0}; 48 if (l==r) 49 { 50 tree[k].m = st[l]; 51 return; 52 } 53 int mid = (l+r)/2; 54 build_tree(l,mid,2*k); 55 build_tree(mid+1,r,2*k+1); 56 tree[k].m = min(tree[2*k].m,tree[2*k+1].m); 57 } 58 59 int inqy(int l,int r,int k) 60 { 61 if (l==tree[k].l&&r==tree[k].r) 62 return tree[k].m; 63 64 int mid = (tree[k].l+tree[k].r)/2; 65 if (r<=mid) return inqy(l,r,2*k); 66 else if (l>mid) return inqy(l,r,2*k+1); 67 return min (inqy(l,mid,2*k) , inqy(mid+1,r,2*k+1)); 68 } 69 70 int main() 71 { 72 int n,m; 73 while (scanf("%d%d",&n,&m)!=EOF) 74 { 75 scanf("%s",str+1); 76 st[0]=0; 77 for (int i=1;i<=n;i++) 78 st[i] = st[i-1] + (str[i]=='('?1:-1 ); 79 build_tree(1,n,1); 80 while (m--) 81 { 82 int a = scan(); 83 int b = scan(); 84 if (a>b) swap(a,b); 85 86 int ok=1; 87 88 int sb = inqy(a,b-1,1); 89 if (str[a]=='(') sb--; 90 else sb++; 91 if (str[b]==')') sb--; 92 else sb++; 93 if (sb<0) ok=0; 94 95 if (ok) 96 printf("Yes "); 97 else 98 printf("No "); 99 } 100 } 101 return 0; 102 }