POJ_1068_Parencodings

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22987 Accepted: 13469

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

Source

Tehran 2001

括号匹配的一个简单的模拟问题

#include <iostream>#include <stdio.h>using namespace std;const int M=22;int pa[2*M];int isu[2*M];void clearu(){    for(int i=0;i<2*M;i++)        isu[i]=0;}int main(){    int t;    int n,num;    int p;    int pl;    scanf("%d",&t);    while(t--)    {        pl=0;p=0;        clearu();        scanf("%d",&n);        for(int j=0;j<n;j++)        {            scanf("%d",&num);            if(num>pl)            {                int tmp=p+num-pl;                pl=num;                for(;p<tmp;p++)                    pa[p]=1;                pa[p]=2;                p++;            }            else if(num==pl)            {                pa[p]=2;                p++;            }        }        for(int i=0;i<n-1;i++)        {            p=0;pl=0;            while(isu[p]||pa[p]==1)                p++;            isu[p]=1;            while(isu[p]||pa[p]==2)            {                p--;                if(isu[p]&&pa[p]!=2)                    pl++;            }            isu[p]=1;            printf("%d ",pl+1);        }        p=0;pl=0;        while(isu[p]||pa[p]==1)            p++;        isu[p]=1;        while(isu[p]||pa[p]==2)        {            p--;            if(isu[p]&&pa[p]!=2)                pl++;        }        isu[p]=1;        printf("%d\n",pl+1);    }    return 0;}