POJ 1068 Parencodings(暴力枚举)

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22856 Accepted: 13399

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

题目大意：对于一个由左括号'（'和右括号'）'组成的字符串S，给出其P序列，输出对应的W序列（P序列中P[i]表示在第i个右括号有P[i]个左括号，W序列中W[i]表示由第i个右括号组成的括号对内有W[i]-1对括号，详见样例）。
基本思路是先由P序列还原出每个右括号在S串中的位置。最后循环每次从后往前扫描到与第i个右括号对应的左括号，输出区间内的左括号数。因为n最大为20，所以暴力也不会超时。

#include<stack>#include<queue>#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int MAXN=1e5+20;int n,r[MAXN],num[MAXN],is[MAXN],vis[MAXN];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif // ONLINE_JUDGE    int tcase;    scanf("%d",&tcase);    while(tcase--)    {        memset(r,0,sizeof(r));        memset(num,0,sizeof(num));        memset(is,0,sizeof(is));        memset(vis,0,sizeof(vis));        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&num[i]);            if(i==0)                r[i]=num[i]+1;            else                r[i]=r[i-1]+num[i]-num[i-1]+1;            is[r[i]]=1;        }        //for(int i=0;i<n;i++)        //    printf("%d ",r[i]);        //puts("");        //for(int i=1;i<=2*n;i++)        //    printf("%s",is[i]?" ）":"（");        //puts("");        for(int i=0;i<n;i++)        {            int cnt=0;            for(int j=r[i]-1;j>=1;j--)            {                if(!is[j])                    cnt++;                else                    continue;                if(vis[j])                    continue;                vis[j]=1;                break;            }            if(i)                printf(" ");            printf("%d",cnt);        }        puts("");    }    return 0;}