UVA 673 栈的应用

You are given a string consisting of parentheses () and []. 

A string of this type is said to be correct:

(a) if it is the empty string

(b) if A and B are correct, AB is correct,

(c) if A is correct, (A) and [A] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. 

Your program can assume that the maximum string length is 128.

Input

The file contains a positive integer n and a sequence of n strings of parentheses ‘()’ and ‘[]’, one stringa line.OutputA sequence of ‘Yes’ or ‘No’ on the output file.

Sample Input

3

([])

(([()])))

([()[]()])()

Sample Output

Yes

No

Yes

题意 很好理解 

1 如果  是空 则合法。

2 如果 A与B合法 ，则AB合法。

3 如果A合法 （A）[A]合法。

用栈来模拟这个过程。

每次取一个元素

如果与栈顶匹配，就pop栈顶，如果不匹配，压入栈中

到最后，如果为空，则合法

错误记录

1.RE

 把 stack < >q定义在了主函数内。运行内存爆了

2.WA

 未考虑情况(1) 即为空行的情况

 错误代码

#include<cstdio>#include<cstring>#include<cstdlib>#include <map>#include <set>#include <vector>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <sstream>#include <string>#define maxn 100000000using namespace std;char s[maxn];stack <char> q;int main(){    int t;    scanf("%d",&t);    while(t--){        while(!q.empty()) q.pop();        cin>>s;        int flag=0;        for(int i = 0 ; s[i] ; i++){            if(s[i]=='('||s[i]=='[')                q.push(s[i]);            else{                if(!q.empty()&&((s[i]==')'&&q.top()=='(')||(s[i]==']'&&q.top()=='[')))                    q.pop();                else{                    flag=1;                    break;                }            }        }        if(!q.empty()) flag = 1;        if(flag) printf("No\n");        else printf("Yes\n");    }}

经过修改后

#include<cstdio>#include<cstring>#include<cstdlib>#include <map>#include <set>#include <vector>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <sstream>#include <string>#define maxn 100000000using namespace std;string s;stack <char> q;int main(){    int t;    //freopen("c:\\MyDrivers\\in.txt","r",stdin);    //freopen("c:\\out.txt","w",stdout);    scanf("%d",&t);    getchar();    while(t--){        while(!q.empty()) q.pop();        getline(cin,s);        int flag=0;        for(int i = 0 ; s[i] ; i++){            if(s[i]=='('||s[i]=='[')                q.push(s[i]);            else{                if(!q.empty()&&((s[i]==')'&&q.top()=='(')||(s[i]==']'&&q.top()=='[')))                    q.pop();                else{                    flag=1;                    break;                }            }        }        if(!q.empty()) flag = 1;        if(flag) printf("No\n");        else printf("Yes\n");    }}

通过 getline 函数来实现 100ms

简化

#include<cstdio>#include<cstring>#include<cstdlib>#include <map>#include <set>#include <vector>#include <iostream>#include <algorithm>#include <queue>#include <stack>#include <sstream>#include <string>#define maxn 100000000using namespace std;stack <char> q;int main(){    int t;    //freopen("f:\\in.txt","r",stdin);    //freopen("f:\\out.txt","w",stdout);    scanf("%d%*c",&t);    while(t--){        while(!q.empty()) q.pop();        char s;        int flag=0;        while((s=getchar()) != '\n'){            if(s=='('||s=='[')                q.push(s);            else{                if(q.empty())                q.push(s);                else if(s==']'&&q.top()=='[')                q.pop();                else if(s==')'&&q.top()=='(')                q.pop();                else q.push(s);            }        }        if(!q.empty()) flag = 1;        if(flag) printf("No\n");        else printf("Yes\n");    }}

通过 getchar（）来读取 高效而便于编程， 时间 20ms

事实上大量输入的情况下 cin 还是挺慢的