POJ 1068 Parencodings

时间：2016-09-03 22:33:04 阅读：157 评论：0 收藏：0 [点我收藏+]

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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;
}

POJ 1068 Parencodings

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原文地址：http://www.cnblogs.com/inmoonlight/p/5837971.html

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