CSU 1809 - Parenthesis - [前缀和+维护区间最小值][线段树/RMQ]
题目链接:http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1809
Bobo has a balanced parenthesis sequence P=p 1 p 2…p n of length n and q questions.
The i-th question is whether P remains balanced after p ai and p bi 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≤10 5,1≤q≤10 5).
The second line contains n characters p 1 p 2…p n.
The i-th of the last q lines contains 2 integers a i,b i (1≤a i,b i≤n,a i≠b i).
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
题意:
现在给出长度为n的一串圆括号串,保证是平衡的括号串;
再给出q个查询,每个查询有两个值a,b,代表询问交换P[a]和P[b],会不会使得括号串变得不平衡。
题解:
首先,我们定义一个preSum数组,代表了这括号序列的前缀和,遇到一个'('就加上1,遇到一个')'就减去1;
这样一来,该括号序列为平衡序列 $\Leftrightarrow$ preSum[n]==0,且对于$\forall$i=1~n都有preSum[i]≥0;
那么我们有以下三种情况:
- P[a]==P[b],这样情况显然交换一下和原来没有区别,显然是保持平衡的;
- P[a]=')'且P[b]='(',这种情况下,preSum[1]~preSum[a-1]不会有任何变动,preSum[a]~preSum[b-1]都会$+=2$,preSum[b]~preSum[n]也不会有任何变动,这样一来,交换后产生的新的括号序列,依然满足preSum[n]==0,且对于$\forall$i=1~n都有preSum[i]≥0,那么显然该括号序列还是平衡序列;
- P[a]='('且P[b]=')',这种情况下,preSum[1]~preSum[a-1]不会有任何变动,preSum[a]~preSum[b-1]都会$-=2$,preSum[b]~preSum[n]不会有任何变动,那么显然关键就在preSum[a]~preSum[b-1]上了,我们知道一旦有一个preSum[i]<0,这个括号序列就不平衡了,所以我们必须保证preSum[a]~preSum[b-1]都大于等于2才行。
所以,我们可以使用线段树或者RMQ维护preSum[]数组的区间最小值,对每次查询只要查询出[a,b]区间内preSum[i]的最小值有没有比2大就可以了。
AC代码:
①线段树:
#include<cstdio> #include<cstring> #include<vector> using namespace std; typedef long long LL; const int maxn=1e5+10; const int INF=0x3f3f3f3f; int n,q; char str[maxn]; int preSum[maxn]; struct Node{ int l,r; int val; }node[4*maxn]; void pushup(int root) { node[root].val=min(node[root*2].val,node[root*2+1].val); } void build(int root,int l,int r) { node[root].l=l; node[root].r=r; if(l==r) node[root].val=preSum[l]; else { int mid=l+(r-l)/2; build(root*2,l,mid); build(root*2+1,mid+1,r); pushup(root); } } int query(int root,int st,int ed) { if(ed<node[root].l || node[root].r<st) return INF; if(st<=node[root].l && node[root].r<=ed) return node[root].val; else return min(query(root*2,st,ed),query(root*2+1,st,ed)); } int main() { while(scanf("%d%d",&n,&q)!=EOF) { scanf("%s",str+1); preSum[0]=0; for(int i=1;i<=n;i++) preSum[i]=preSum[i-1]+(str[i]=='('?1:-1); build(1,1,n); for(int i=1,a,b;i<=q;i++) { scanf("%d%d",&a,&b); if(a>b) swap(a,b); if( str[a]==str[b] || (str[a]==')' && str[b]=='(') ) printf("Yes\n"); else if(query(1,a,b-1)>=2) printf("Yes\n"); else printf("No\n"); } } }
②RMQ:
#include<cstdio> #include<cstring> #include<vector> #include<cmath> using namespace std; typedef long long LL; const int maxn=1e5+10; const int INF=0x3f3f3f3f; int n,q; char str[maxn]; int preSum[maxn]; struct _RMQ{ int Mnum[maxn][20]; //int(log(maxn)/log(2.0)) void init(int num[]) { for(int i=1;i<=n;i++) Mnum[i][0]=num[i]; int j_max=(log(n)/log(2)); for(int j=1;j<=j_max;j++) { for(int i=1;i<=n;i++) { if(i+(1<<j)-1 <= n) Mnum[i][j]=min(Mnum[i][j-1],Mnum[i+(1<<(j-1))][j-1]); } } } int query(int l,int r) { int k=log(r-l+1)/log(2); return min(Mnum[l][k],Mnum[r-(1<<k)+1][k]); } }RMQ; int main() { while(scanf("%d%d",&n,&q)!=EOF) { scanf("%s",str+1); preSum[0]=0; for(int i=1;i<=n;i++) preSum[i]=preSum[i-1]+(str[i]=='('?1:-1); RMQ.init(preSum); for(int i=1,a,b;i<=q;i++) { scanf("%d%d",&a,&b); if(a>b) swap(a,b); if( str[a]==str[b] || (str[a]==')' && str[b]=='(') ) printf("Yes\n"); else if(RMQ.query(a,b-1)>=2) printf("Yes\n"); else printf("No\n"); } } }
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