poj 1068（模拟题）

时间：2015-12-16 23:00:20 阅读：316 评论：0 收藏：0 [点我收藏+]

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23545		Accepted: 13802

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

思路：简单回溯即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define PI 3.141592653589792128462643383279502
int main(){
    freopen("in.txt","r",stdin);
    int t,n,m,c,p[30],pp[30];
    bool w[50];
    cin>>t;
    while(t--){
        memset(w,true,sizeof(w));
        cin>>n;
        c=1;
        for(int i=1;i<=n;i++){
            cin>>m;
            w[m+i]=false;
            p[c++]=m+i;
        }
        int k=0,b=0,kk;
        for(int i=1;i<c;i++){
            kk=p[i];
            for(int j=kk;j>=1;j--){
                if(w[j]==false)
                    k++;
                else
                    b++;
                if(b-k==0) {
                    pp[i]=k;
                    k=0;b=0;
                    break;
                }
            }
        }
        for(int i=1;i<c;i++){
            if(i!=1)cout<<" ";
            cout<<pp[i];
        }
        cout <<endl;
    }
    return 0;
}

poj 1068（模拟题）

原文：http://www.cnblogs.com/yoyo-sincerely/p/5052494.html

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