試題分析 mission 元素 第一個 href arc etc space ans

[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 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