POJ 1068 Parencodings -YY题

POJ 1068 Parencodings ---YY题

Language:DefaultParencodingsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 15029 Accepted: 8959

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

给你一组数

每个数字表示每个右括号之前有几个左括号

要你找出每个 右括号匹配的左括号 从这个右括号向左数开始是第几个左括号...(真拗口...)

其实看看样例就明白了...

思路就是还原括号...构建一个数组储存括号

再遍历一遍找到匹配的括号

#include <set>#include <list>#include <cmath>#include <ctime>#include <deque>#include <queue>#include <stack>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <cassert>#include <cstdlib>#include <cstring>#include <sstream>#include <iostream>#include <algorithm>using namespace std;int seq[500],a[30],b[500],x,n,i,j,k,cnt;                                                         int main(){//freopen("in.txt","r",stdin);cin>>x;while(x--){memset(b,0,sizeof(b));cin>>n;for(i=1,j=0;i<=n;i++){cin>>a[i];for(k=0;k<a[i]-a[i-1];k++,j++)seq[j]=0;seq[j++]=1;}for(i=0;i<j;i++){if(seq[i]==1){for(k=i-1,cnt=1;k>=0;k--){if(seq[k]==0 && b[k]==1)cnt++;else if(seq[k]==0 && b[k]==0){b[k]=1;cout<<cnt<<" ";break;}}}}cout<<endl;}    return 0;}

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