标签:poj bin string idt 复杂度 结果 pos ++ 输出
[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
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
标签:poj bin string idt 复杂度 结果 pos ++ 输出
原文地址:http://www.cnblogs.com/wxjor/p/6974129.html