hdu 5831 Rikka with Parenthesis II 线段树
2016-08-11 22:51
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Rikka with Parenthesis II
题目连接:
http://acm.hdu.edu.cn/showproblem.php?pid=5831Description
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:Correct parentheses sequences can be defined recursively as follows:
1.The empty string "" is a correct sequence.
2.If "X" and "Y" are correct sequences, then "XY" (the concatenation of X and Y) is a correct sequence.
3.If "X" is a correct sequence, then "(X)" is a correct sequence.
Each correct parentheses sequence can be derived using the above rules.
Examples of correct parentheses sequences include "", "()", "()()()", "(()())", and "(((())))".
Now Yuta has a parentheses sequence S, and he wants Rikka to choose two different position i,j and swap Si,Sj.
Rikka likes correct parentheses sequence. So she wants to know if she can change S to a correct parentheses sequence after this operation.
It is too difficult for Rikka. Can you help her?
Input
The first line contains a number t(1<=t<=1000), the number of the testcases. And there are no more then 10 testcases with n>100For each testcase, the first line contains an integers n(1<=n<=100000), the length of S. And the second line contains a string of length S which only contains ‘(’ and ‘)’.
Output
For each testcase, print "Yes" or "No" in a line.Sample Input
34
())(
4
()()
6
)))(((
Sample Output
YesYes
No
Hint
题意
给你一个括号序列,你必须交换俩括号位置,问你可以可以使他合法。题解:
哎呀,好气啊,感觉只有我们队是用线段树去模拟交换的……最优情况下一定交换第一个右括号和最后一个左括号,交换后判断一下即可。 时间复杂度 O(n)O(n)
代码
#include <bits/stdc++.h> #define rep(a,b,c) for(int (a)=(b);(a)<=(c);++(a)) #define drep(a,b,c) for(int (a)=(b);(a)>=(c);--(a)) #define pb push_back #define mp make_pair #define sf scanf #define pf printf #define two(x) (1<<(x)) #define clr(x,y) memset((x),(y),sizeof((x))) #define dbg(x) cout << #x << "=" << x << endl; const int mod = 1e9 + 7; int mul(int x,int y){return 1LL*x*y%mod;} int qpow(int x , int y){int res=1;while(y){if(y&1) res=mul(res,x) ; y>>=1 ; x=mul(x,x);} return res;} inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;} using namespace std; const int maxn = 100000 + 150; int len,prefix[maxn],lstid; char str[maxn]; struct Sgtree{ struct node{ int l , r , lzy , mi ; void Update( int v ){ lzy += v; mi += v; } }tree[maxn << 2]; void Build( int l , int r , int o ){ tree[o].l = l , tree[o].r = r , tree[o].lzy = tree[o].mi = 0; if( r > l ){ int mid = l + r >> 1; Build( l , mid , o << 1 ); Build( mid + 1 , r , o << 1 | 1 ); Maintain( o ); }else{ if( l == len ) lstid = o; tree[o].mi = prefix[l]; } } int query( int x , int o ){ int l = tree[o].l , r = tree[o].r; if( l == r ) return tree[o].mi; else{ int mid = l + r >> 1 , rs ; ReleaseLabel(o); if( x <= mid ) rs = query( x , o << 1 ); else rs = query( x , o << 1 | 1 ); Maintain(o); return rs; } } void Maintain( int o ){ tree[o].mi = min( tree[o << 1].mi , tree[o << 1 | 1 ].mi ); } void ReleaseLabel( int o ){ if( tree[o].lzy ){ tree[o << 1].Update( tree[o].lzy ); tree[o << 1 | 1].Update( tree[o].lzy ); } tree[o].lzy = 0; } void Modify( int ql , int qr , int y , int o ){ int l = tree[o].l , r = tree[o].r; if( ql <= l && r <= qr ) tree[o].Update( y ); else{ int mid = l + r >> 1; ReleaseLabel( o ); if( ql <= mid ) Modify( ql , qr , y , o << 1 ); if( qr > mid ) Modify( ql , qr , y , o << 1 | 1 ); Maintain( o ); } } int Search_First( int x , int o ){ int l = tree[o].l , r = tree[o].r; if( l == r ){ if( tree[o].mi < 0 ) return l; return -1; } int mid = l + r >> 1 , rs = -1; ReleaseLabel( o ); if( tree[o << 1].mi < 0 ) rs = Search_First( x , o << 1 ); if( rs == -1 && x > mid && tree[o << 1 | 1].mi < 0 ) rs = Search_First( x , o << 1 | 1 ); Maintain( o ); return rs; } }Sgtree; bool solve(){ rep(i,1,len) if(str[i]=='(' && i > 1){ int go = Sgtree.Search_First( i - 1 , 1 ); if( go != -1 ){ Sgtree.Modify( go , i - 1 , 2 , 1 ); if( Sgtree.tree[1].mi == 0 && Sgtree.query(len,1) == 0 ) return true; Sgtree.Modify( go , i - 1 , -2 , 1 ); } } return false; } int main(int argc,char *argv[]){ int T=read(); while(T--){ len=read(); sf("%s",str+1); int ar = 0 , ok = 1; rep(i,1,len){ prefix[i] = prefix[i - 1]; if(str[i]=='('){ ++ prefix[i]; ++ ar; } else{ if( ar == 0 ) ok = 0; -- ar; -- prefix[i]; } } Sgtree.Build(1,len,1); if( ar != 0 ) ok = 0; if( ok ){ if( len == 2 ) pf("No\n"); else pf("Yes\n"); }else{ bool result = solve(); if( result == true ) printf("Yes\n"); else printf("No\n"); } } return 0; }
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