POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24932		Accepted: 14695

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

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

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

Source

Tehran 2001

解析：模拟。

#include <cstdio>

char s[10000000];

int a[25];

int n;

int res[25];

int rid[25];

void solve()

{

    int cnt = 0, rcnt = 0;

    for(int i = 1; i <= a[0]; ++i)

        s[cnt++] = '(';

    s[cnt] = ')';

    rid[rcnt++] = cnt;

    ++cnt;

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

        int num = a[i]-a[i-1];

        for(int j = 1; j <= num; ++j)

            s[cnt++] = '(';

        s[cnt] = ')';

        rid[rcnt++] = cnt;

        ++cnt;

    }

    int res_cnt = 0;

    for(int i = 0; i < rcnt; ++i){

        int l = 0, r = 1;

        for(int j = rid[i]-1; ; --j){

            if(s[j] == '('){

                ++l;

                if(l == r){

                    res[res_cnt++] = l;

                    break;

                }

            }

            else

                ++r;

        }

    }

    for(int i = 0; i < n-1; ++i)

        printf("%d ", res[i]);

    printf("%d\n", res[n-1]);

}

int main()

{

    int t;

    scanf("%d", &t);

    while(t--){

        scanf("%d", &n);

        for(int i = 0; i < n; ++i)

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

        solve();

    }

    return 0;

}

巴特西

POJ 1068 Parencodings

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