【数据结构(高效)/暴力】Parencodings

[poj1068] Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26686		Accepted: 15645

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

Source

Tehran 2001

 

试题分析：这题标程是暴力，但我做题喜欢乱搞，时间复杂度就自然没有暴力那么高:O(TN)，可惜没有给O(TN)与暴力分开 TAT

                那么这题O(TN)如何做呢？我们从左到右瞟一眼这个序列，发现可以用两个栈(一个记录括号数，一个记录提供括号的编号，相信往下读会更有体会)，功能如下(以样例2为例)：

                ①遇到4，第一个右括号左边有4个左括号，它自己匹配到一个，还剩3个给后面的用，将3入栈

                ②遇到6，跟先前不一样的话直接输出1，将(6-4)-1个左括号入栈

                ③又遇到6，发现前面没有了，从栈中弹出一个括号(即将栈顶元素弹出，括号数-1后在塞回去，0不要塞)，把它的对应编号弹出，累积结果，再压回编号的栈中

                ④又遇到6，继续上面的操作

                ⑤遇到8，输出1，把剩余括号压入栈中，另一个栈记录编号

                ⑥遇到9，跟前面差1，这个1是给9自己用的，不能压栈

 

                整个算法过程就是这样，用栈O(TN)，数组模拟也可以

 

代码

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<stack>
#include<algorithm>
using namespace std;
inline int read(){
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-'0';
    return x*f;       
}
int N;
int P[1001];
int ans[1001];
stack<int> sta;
stack<int> k;
int main(){
    int T=read();
    while(T--){
        stack<int> sta;
        stack<int> k;
        N=read();
        for(int i=1;i<=N;i++){
            P[i]=read();
            if(P[i]!=P[i-1]){
               ans[i]=1;
               if(P[i]-P[i-1]!=1) sta.push(P[i]-P[i-1]-1),k.push(i);
            }
            else{
                int s=sta.top();
                sta.pop();
                ans[i]=i-k.top()+1;
                if(s-1!=0) sta.push(s-1);
                else k.pop();
            }
        }
        for(int i=1;i<N;i++)
            printf("%d ",ans[i]);
        printf("%d
",ans[N]);
    }
}