POJ 1068 Parencodings 栈模拟

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

思路：模拟

代码：

#include <stdio.h>
#include <string.h>

int t, n, first[255], i, trans[255], ans[255], now, stack[255], num[255], top, ansn;

void tra() {
	int l = 0, i, j, r = 0;
	for (i = 1; i <= n; i ++) {
		for (j = 0; j < first[i] - l; j ++) {
			trans[now ++] = 1;
		}
		trans[now ++] = 2;
		r ++;
		l = first[i];
	}
	for (i = 0; i < l - r; i ++)
		trans[now ++] = 2;
}

void tra2() {
	int i, j;
	for (i = 1; i < now; i ++) {
		if (trans[i] == 1) {
			stack[top ++] = trans[i];
		}
		if (trans[i] == 2) {
			top --;
			ans[ansn ++] = num[top];
			num[top] = 0;
			for (j = 0; j <= top - 1; j ++)
				num[j] ++;
		}
	}
}
int main() {
	scanf("%d", &t);
	while (t --) {
		now = 1; top = 1; ansn = 1;
		memset(stack, 0, sizeof(stack));
		memset(num, 0, sizeof(num));
		memset(trans, 0, sizeof(trans));
		memset(ans, 0, sizeof(ans));
		scanf("%d", &n);
		for (i = 1; i <= n; i ++)
			scanf("%d", &first[i]);
		tra();
		tra2();
		for (i = 1; i < ansn - 1; i ++)
			printf("%d ", ans[i] + 1);
		printf("%d\n", ans[ansn - 1] + 1);
	}
	return 0;
}