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POJ 1068 Parencodings

2024-08-11 22:41:56   218人阅读

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

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
 
 
 
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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