POJ 1068 Parencodings <模拟>

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19352 Accepted: 11675

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

题目大意：p代表，每一个右括号前有几个左括号，w代表每一个右括号和他前面的第几个括号匹配。

我们可以直接用数组来存，1代表左括号，0代表右括号，然后循环去找每一个右括号出现的位置前的左括号，如果这个左括号被使用过了

就标记为-1，下一个右括号匹配到这的时候步数加一即可

#include <iostream>#include <cstring>#include <stack>#include <cstdio>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <set>#include <map>const double eps=1e-8;const double PI=acos(-1.0);using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,a,a1=0;        int s[50];//数字1代表左括号，0代表右括号;        scanf("%d",&n);        int j=0;        for(int i=0; i<n; i++)        {            scanf("%d",&a);            int tt=a-a1;            while(tt--)            {                s[j++]=1;            }            a1=a;            s[j++]=0;        }        int flag=0;       /*  for(int i=0; i<=2*n-1; i++)            cout<<s[i]<<" ";         cout<<endl;*/        for(int i=0; i<=2*n-1; i++)        {            if(s[i]==0)            {                int cou=1;                for(int k=i-1; k>=0; k--)                {                    if(s[k]==0)continue;                    if(s[k]==-1) cou++;                    else if(s[k]==1)                    {                        if(!flag)                        {                            printf("%d",cou);                            flag=1;                        }                        else                            printf(" %d",cou);                        s[k]=-1;                        break;                    }                }            }        }        printf("\n");    }    return 0;}