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2016年湖南省第十二届大学生计算机程序设计竞赛---Parenthesis(线段树求区间最值)

原题链接

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1809

 

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 32 32 1()1 2

Sample Output

NoYesNo

Source

湖南省第十二届大学生计算机程序设计竞赛

 

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]  s[b]交换后是否任然平衡,平衡则输出“Yes”  否则输出“No”;

思路:定义数组num[] ,num[i]表示s[1]到s[i]中左括号数减去右括号数的差值,分析可知因为s是平衡括号序列,那么num[i]>=0   令a<b  ,那么交换s[a] s[b]后,只对num[a]~num[b-1]产生影响,并且交换后当num[k]<0(a<=k<b)时表示不平衡,而只有s[a]=‘(‘  s[b]=‘)‘ 交换后才可能使num[k]<0 。所以特判当s[a]=‘(‘  s[b]=‘)‘时,用线段树求区间a~b-1的最小值,当最小值小于2时,即交换后不平衡,为什么呢?   s[a]=‘(‘  s[b]=‘)‘交换后num[a]~num[b-1]都减2;

代码如下:

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#define eps 1e-8#define maxn 105#define inf 0x3f3f3f3f3f3f3f3f#define IN freopen("in.txt","r",stdin);using namespace std;char s[100005];int num[100005];int a,b; struct Node{    //int l,r;    int v;}node[100005*4]; void build(int l,int r,int i){    if(l==r) {       node[i].v=num[l];       return;    }    int mid=(l+r)>>1;    build(l,mid,i<<1);    build(mid+1,r,i<<1|1);    node[i].v=min(node[i<<1].v,node[i<<1|1].v);} void query(int l,int r,int &tmp,int i){    if(l>=a&&r<=b) { tmp=node[i].v; return; }    int mid=(l+r)>>1;    if(mid>=b) query(l,mid,tmp,i<<1);    else  if(mid<a) query(mid+1,r,tmp,i<<1|1);    else {          int tmp2;          query(l,mid,tmp,i<<1);          query(mid+1,r,tmp2,i<<1|1);          tmp=min(tmp,tmp2);    }} int main(){    int n,q;    while(scanf("%d%d",&n,&q)!=EOF)    {        scanf("%s",s+1);        num[0]=0;        for(int i=1;i<=n;i++)        {            num[i]=num[i-1];            if(s[i]==() num[i]++;            else  num[i]--;        }        build(1,n,1);        while(q--)        {            scanf("%d%d",&a,&b);            if(a>b) swap(a,b);             if(s[a]==s[b]||s[a]==)&&s[b]==()  puts("Yes");            else            {                b--;                int tmp=99999999;                query(1,n,tmp,1);                if(tmp<2) puts("No");                else      puts("Yes");            }        }    }    return 0;}/**8 8(())(())2 7*/

 

2016年湖南省第十二届大学生计算机程序设计竞赛---Parenthesis(线段树求区间最值)