POJ 1068 Parencodings 模拟 栈操作

Parencodings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17398 Accepted: 10478

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

这道题的确英语有丝绕，看的纠结，但是大体意思就是有一串括号，序列P代表每一个右括号左面有多少个左括号，序列序列W代表每一个右括号左面第几个左括号是与之配对的左括号，然后给定P序列求W序列，这道题其实看到括号配对就要想到栈操作，先根据P序列还原原来的括号序列，主页君用了一个标记方法方便后面操作，还原后，正数代表左括号，负数代表右括号，数字代表其是第几个数，之后，就是栈操作，对每个数字判断，如果是正数，即左括号，则入栈，如果是负数，即右括号，此时将负数的变成相反的正数减去栈顶的数字除以2即为中间有多少对括号配对，也代表中间相隔多少个左括号，再+1就是右括号配对左括号是第几个左括号，将此序列再存入一个数组，待所有括号处理结束后，输出此数组即可AC。。。其实题目还是挺简单的。

下面是AC代码：

#include<cstdio>#include<iostream>#include<stack>using namespace std;int a[25],b[25],c[100];int main(){int i,j,t,n,p,num;stack<int> st;scanf("%d",&t);while(t--){num=0;cin>>n;a[0]=0;p=0;for(i=1;i<=n;i++){cin>>a[i];for(j=p;j<p+a[i]-a[i-1];j++)c[j]=j+1;c[p+a[i]-a[i-1]]=-(p+a[i]-a[i-1]+1);p=p+a[i]-a[i-1]+1;}j=0;for(i=0;i<p;i++){if(c[i]>0)st.push(c[i]);else{num=st.top();st.pop();num=(-c[i]-num)/2+1;b[j]=num;j++;}}for(i=0;i<j-1;i++)printf("%d ",b[i]);printf("%d\n",b[j-1]);}return 0;}