模拟 --- 简单括号匹配

Parencodings

Description

Let S = s1 s2...s2n be a well-formed string of
parentheses. S can be encoded in two different ways:
q By an integer
sequence P = p1 p2...pn where pi is the number of left parentheses before the
ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1
w2...wn where for each right parenthesis, say a in S, we associate an integer
which is the number of right parentheses counting from the matched left
parenthesis of a up to a. (W-sequence).

Following is an example of the
above encodings:

	S		(((()()())))


	P-sequence	    4 5 6666


	W-sequence	    1 1 1456

Input

The first line of the input contains a single
integer t (1 <= t <= 10), the number of test cases, followed by the input
data for each test case. The first line of each test case is an integer n (1
<= n <= 20), and the second line is the P-sequence of a well-formed
string. It contains n positive integers, separated with blanks, representing
the P-sequence.

Output

The output file consists of exactly t lines
corresponding to test cases. For each test case, the output line should contain
n integers describing the W-sequence of the string corresponding to its given
P-sequence.

Sample Input

2

6

4 5 6 6 6 6

9

4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 6

1 1 2 4 5 1 1 3 9

【题目来源】

Tehran
2001

http://poj.org/problem?id=1068

【题目大意】

给你一串数字，每个数字的位置都是一个右括号，并且本身的值表示该右括号的左边有多少左括号。

现在要你匹配这些右括号，输出一列数字，这些数字表示该右括号到达与该右括号匹配的左括号这一段中有多少个左括号。

翻译得太差强人意了，有时候题意只可意会，不可言传==||

【题目分析】

题目很简单，直接模拟就可。

AC代码：

#include<cstdio>

#include<cstdlib>

#include<iostream>

#include<cstring>

#include<algorithm>

#define MAX 100000

using namespace std;

struct Node

{

    bool lor;

    bool vis;

};

Node node[100000];

int main()

{

    int T;

    cin>>T;

    while(T--)

    {

        int n;

        int i,j;

        int a[25];

        for(i=0;i<MAX;i++)

            node[i].vis=false;

        cin>>n;

        a[0]=0;

        for(i=1;i<=n;i++)

        {

            scanf("%d",&a[i]);

        }

        int index=-1;

        int temp;

        for(i=1;i<=n;i++)

        {

            temp=a[i]-a[i-1];

            for(j=0;j<temp;j++)

            {

                node[++index].lor=0;

            }

            node[++index].lor=1;

        }

//        cout<<index<<endl;

//        for(i=0;i<=index;i++)

//            printf("%d ",node[i].lor);

//        puts("");

        int cnt;

        int ans[MAX];

        int index1=-1;

        for(i=0;i<=index;i++)

        {

            cnt=0;

            if(node[i].lor==1)

            {

                for(j=i;j>=0;j--)  //倒过来数

                {

                    if(node[j].lor==0)

                    {

                        cnt++;

                       if(!node[j].vis)

                            {

                                node[j].vis=true;

                                ans[++index1]=cnt;

//                                cout<<"cnt="<<cnt<<endl;

                                break;

                            }

                    }

                }

            }

            cnt=0;

        }

//        cout<<"index1="<<index1<<endl;

        for(i=0;i<=index1;i++)

        {

            if(i==0)

               cout<<ans[i];

            else

               cout<<" "<<ans[i];

        }

        puts("");

    }

    return 0;

}

模拟 --- 简单括号匹配