【數據結構(高效)/暴力】Parencodings
阿新 • • 發佈:2017-06-10
試題分析 mission 元素 第一個 href arc etc space ans [poj1068] Parencodings
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:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
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).
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
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 試題分析:這題標程是暴力,但我做題喜歡亂搞,時間復雜度就自然沒有暴力那麽高:O(TN),可惜沒有給O(TN)與暴力分開 TAT 那麽這題O(TN)如何做呢?我們從左到右瞟一眼這個序列,發現可以用兩個棧(一個記錄括號數,一個記錄提供括號的編號,相信往下讀會更有體會),功能如下(以樣例2為例): ①遇到4,第一個右括號左邊有4個左括號,它自己匹配到一個,還剩3個給後面的用,將3入棧 ②遇到6,跟先前不一樣的話直接輸出1,將(6-4)-1個左括號入棧 ③又遇到6,發現前面沒有了,從棧中彈出一個括號(即將棧頂元素彈出,括號數-1後在塞回去,0不要塞),把它的對應編號彈出,累積結果,再壓回編號的棧中 ④又遇到6,繼續上面的操作 ⑤遇到8,輸出1,把剩余括號壓入棧中,另一個棧記錄編號 ⑥遇到9,跟前面差1,這個1是給9自己用的,不能壓棧 整個算法過程就是這樣,用棧O(TN),數組模擬也可以 代碼#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]); } }
【數據結構(高效)/暴力】Parencodings