Winter-2-STL-B Brackets 解题报告及测试数据

Time Limit:2000MS     Memory Limit:65536KB

Description

Given a string consisting of brackets of two types find its longest substring that is a regular brackets sequence.

Input

There are mutiple cases in the input file.

Each case contains a string containing only characters ‘(’ , ‘)’ , ‘[’ and ‘]’ . The length of the string does not exceed 100,000.

There is an empty line after each case.

Output

Output the longest substring of the given string that is a regular brackets sequence.

There should be am empty line after each case.

Sample Input

([(][()]]()

([)]

Sample Output

[()]

题解：

这道题是用栈进行括号的匹配过程。具体思路如下：

1、使用字符串str[100005]储存输入的括号，遍历操作。每个字符有三种情况，（1）“(]”或“[)”属于无法匹配，之前栈中的括号无法进行后续匹配，所以进行清空栈。（2）当前遍历到的字符可以与栈顶元素抵消，那么此时用到了一个标记数组p，将栈顶元素和当前元素位置标记为1。（3）不属于前两种情况，进栈等待匹配。由函数 can_place(char)进行判断。

2、将匹配位置都置1后，需要判断全为1最长子串，这很容易，使用st更新起始位置，maxl更新长度即可。

 以下是代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <stack>
#include <cstring>
#include <utility>
using namespace std;
char str[100005];
int p[100005];//标记括号成功匹配
stack<pair<char,int> >v;
int can_place(char ch){//如果非法，返回0，可进栈，返回1，可与栈顶抵消，返回2
    if(v.empty()){
        if(ch == '(' || ch == '[')return 1;
        return 0;
    }else switch(v.top().first){
        case'(':if(ch ==']')return 0;if(ch==')')return 2;return 1;
        case'[':if(ch==')')return 0;if(ch==']')return 2;return 1;
    }
}
int main(){
    //freopen("1.in","r",stdin);
    int len;
    while(scanf("%s",str)!=EOF){
        len = strlen(str);
        memset(p,0,sizeof(p));
        for(int i=0;i<len;i++)
        switch(can_place(str[i])){
            case 0:while(!v.empty())v.pop();break;//遇到非法括号，清空栈
            case 1:v.push(make_pair(str[i],i));break;//若可以进栈，进栈
            case 2:p[i]=p[v.top().second]=1;v.pop();//若匹配出栈，将对应位置置1
        }
        while(!v.empty())v.pop();
        int maxl=0,tl=0,t=0,st=0;
        for(int i=0;i<len;i++){//计算最长的全1子串
            if(!p[i]){
                t=i+1;
                tl=0;
            }else tl++;
            if(maxl<tl){//更新长度maxl及开始位置st
                st=t;
                maxl=tl;
            }
        }
        for(int i=st;i< st+maxl;i++)
            printf("%c",str[i]);
        printf("\n\n");
    }
}

以下是测试数据:

smaple input

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

()()[[](]

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

​sample output

(()[()])

()()

[][()()]