poj1141 Brackets Sequence

黑书p113上的题。

我的思路:

f[i][j]：i-j的最短规则序列长度

f[i][j] = f[i][j - 1] + 2;

当 s[j] == ')' && s[k] == '(' || s[j] == ']' && s[k] == '[' 时

　　f[i][j] <?= f[i][k - 1] + f[k + 1][j];

/**

 * Problem:POJ1141

 * Author:Shun Yao

 * Time:2013.5.19

 * Result:Accepted

 * Memo:DP

 */



#include <cstring>

#include <cstdlib>

#include <cstdio>



using namespace std;



long n, f[105][105], g[105][105];

char s[105];



void print_the_solution(long i, long j) {

	if (i > j)

		return;

	if (!g[i][j]) {

		print_the_solution(i, j - 1);

		if (s[j] == '(' || s[j] == ')')

			printf("()");

		else

			printf("[]");

	} else {

		print_the_solution(i, g[i][j] - 1);

		printf("%c", s[g[i][j]]);

		print_the_solution(g[i][j] + 1, j - 1);

		printf("%c", s[j]);

	}

}



int main() {

	static long i, j, k, l, tmp;

	freopen("poj1141.in", "r", stdin);

	freopen("poj1141.out", "w", stdout);

	scanf("%s", s + 1);

	n = strlen(s + 1);

	for (i = 1; i <= n; ++i) {

		memset(f[i], 0, sizeof f[i]);

		memset(g[i], 0, sizeof g[i]);

		f[i][i] = 2;

	}

	for (l = 2; l <= n; ++l)

		for (i = 1; i <= n - l + 1; ++i) {

			j = i + l - 1;

			f[i][j] = f[i][j - 1];

			if (s[j] == ']') {

				for (k = i; k < j; ++k)

					if (s[k] == '[' && f[i][j] > (tmp = f[i][k - 1] + f[k + 1][j - 1])) {

						f[i][j] = tmp;

						g[i][j] = k;

					}

			} else if (s[j] == ')') {

				for (k = i; k < j; ++k)

					if (s[k] == '(' && f[i][j] > (tmp = f[i][k - 1] + f[k + 1][j - 1])) {

						f[i][j] = tmp;

						g[i][j] = k;

					}

			}

			f[i][j] += 2;

		}

	print_the_solution(1, n);

	putchar('\n');

	fclose(stdin);

	fclose(stdout);

	return 0;

}