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【数据结构(高效)/暴力】Parencodings

2024-10-03 16:41:38   218人阅读

[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

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
 
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\n",ans[N]);    }}

　　

【数据结构(高效)/暴力】Parencodings


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