Parencodings
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Let S = s1 s2...s2n be a well-formed string of parentheses. Scan be encoded in two different ways:
q By an integer sequence P = p1 p2...pn where pi is the number ofleft parentheses before the ith right parenthesis in S(P-sequence).
q By an integer sequence W = w1 w2...wn where for each rightparenthesis, say a in S, we associate an integer which is thenumber of right parentheses counting from the matched leftparenthesis 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 tothe W-sequence of the same string.
q By an integer sequence P = p1 p2...pn where pi is the number ofleft parentheses before the ith right parenthesis in S(P-sequence).
q By an integer sequence W = w1 w2...wn where for each rightparenthesis, say a in S, we associate an integer which is thenumber of right parentheses counting from the matched leftparenthesis 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 tothe W-sequence of the same string.
- Input
The first line ofthe input contains a single integer t (1 <= t <= 10), thenumber of test cases, followed by the input data for each testcase. 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-formedstring. It contains n positive integers, separated with blanks,representing the P-sequence.
- Output
The output fileconsists of exactly t lines corresponding to test cases. For eachtest case, the output line should contain n integers describing theW-sequence of the string corresponding to its givenP-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
题目分析:
此题多组处理数据,每组输入n个数,每个a[i]代表当前对应的右括号前面共有多少个左括号。
比如:n=6, 4 5 6 6 6 6 。 对应第一个右括号前有4个左括号,第二个右括号前有5个左括号,不过千万不要落下第一个右括号啊,以此类推下去。。。。。。
现在要求输出另一个数组,数组元素为当前右括号包含的括号总数!
比如:(()())
第一个右括号只包含自己,故为 1;
第二个右括号也只包含自己, 故为 1;
第三个右括号包含前面两个括号,再加上自己,故为 3.
所以应输出:1 1 3 ;
算法思路:根据n个数的数组模拟出原来的括号字符串,而我用的整形数组代替的,就是用1代表'(',用2代表‘)’,然后对这个整形数组处理。
样例解析:
(((()()()))))
1 1 112 12 12 2 2 2 2
数组找到2了,则往2的前面去找1,也就是所谓的左括号,然后将找到的左括号标记为0,接下来将退出。
若回溯时遇到0,则计数器+1,继续回溯直到找到1.
就是输入部分写得麻烦点。。
#include
using namespace std;
int S[10005];
int main()
{
int t,n;
cin >> t;
while (t--)
{
memset(S, 0, sizeof(S));
int i=0,a=0,temp=0,a0=0;
cin >> n;
while (n--)
{
cin >> a;
temp = a - a0;
for (int cnt = 0; cnt < temp; cnt++)
{
S[i] = 1; //用1/2表示左右括号
i++;
}
S[i] = 2;
i++;
a0 = a;
} //输入同时转化储存部分
int flag = 0;
for (int k = 0; k < i; k++)
{
int j = k;
if (S[j] == 2)
{
int cnt = 1;
do
{
j--;
if (S[j] == 0)
cnt++;
if (S[j] == 1)
{
S[j] = 0;
break;
}
} while (j >= 0);
if (flag)
cout << " " << cnt;
else
cout << cnt;
flag = 1;
}
}
cout << endl;
}
return 0;
}
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