POJ1068 Parencodings(模拟)

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 20763 Accepted: 12480

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 6666W-sequence    1 1 1456

Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.

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

264 5 6 6 6 69 4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 61 1 2 4 5 1 1 3 9

解题思路:

本题为模拟题，为了简单化，可以把括号转化成0和1存储，0代表左括号，1代表右括号，根据推理可得如下公式:location = a[i] + i,location代表右括号的位置，也就是现在1所在的位置，可以转化成0和1的串，最后再用回溯法，把括号进行配对，配对完成的置为访问过，边配对边输出结果，一直到所有的都配对完。在poj上本题用c++提交wrong answer,用g++ Accepted,本题槽点略多，看poj 的discuss就知道了。

代码如下:

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int maxn = 500;const int maxn1 = 500;int a[maxn];int m;void change(){    int b[maxn1];    int vis[maxn1];    int i,j;    int location;    int temp;    int k;    int pos;    int cnt;    memset(b,-1,sizeof(b));    memset(vis,0,sizeof(vis));    /*for(i=0;i<300;i++)    {        b[i] = -1;    }    for(i=0;i<300;i++)    {        vis[i] = 0;    }*/    temp = 0;    cnt = 0;    for(i=0;i<m;i++)    {        location = a[i] + i;   //计算右括号的位置        for(j=temp;j<=location;j++)          {            b[j] = 0;            temp = location + 1;        }        b[location] = 1;    }    k = 0;    while(b[k] != -1)    {        k++; //0,1一共的个数    }    for(i=0;i<k;i++)    {        if(b[i] == 1)        {            for(j=i-1;j>=0;j--)            {                if(b[j] == 0 && vis[j] == 0)                {                    vis[i] = 1;                    vis[j] = 1;                    pos = i-j+1;                    cnt++;                    if(cnt == 1)                        cout<<pos/2;                    else                        cout<<" "<<pos/2;                    break;                }            }        }    }    cout<<endl;}int main(){    int T;    int i;    //freopen("111","r",stdin);    cin>>T;    while(T--)    {        cin>>m;        for(i=0;i<m;i++)        {            cin>>a[i];        }        change();    }    return 0;}