Parencodings(POJ 1068)(模擬)
題目連結:http://poj.org/problem?id=1068
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
思考:純模擬的題不要怕麻煩,按照題中的模擬的步驟做就行了。只要時間複雜度不超就行。
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
int t;
cin >> t;
while (t--) {
int n;
int a[100];
char b[1000];
int vis[1000];
int ne[1000];
memset (a, 0, sizeof(a));
memset (b, '\0', sizeof(b));
memset (vis, 0, sizeof(vis));
memset (ne, 0, sizeof(ne));
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
int num = 0;
int num2 = 0;
for (int i = 1; i <= n; i++) {
for (int j = 0; j < a[i] - a[i - 1]; j++) {
b[num++] = '(';
}
ne[num2++] = num;
b[num++] = ')';
}
for (int i = 0; i < num2; i++) {
int cnt = 0;
for (int j = ne[i]; j >= 0; j--) {
if (b[j] == '(') {
cnt++;
if (!vis[j]) {
vis[j] = 1;
break;
}
}
}
cout << cnt << " ";
}
cout << endl;
}
return 0;
}
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