UVa 348 Optimal Array Multiplication Sequence (区间DP&矩阵链乘,MCM)

348 - Optimal Array Multiplication Sequence

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=284

记忆化搜索：dp[a][b] = max(dp[a][b], dp[a][i] + dp[i + 1][b] + x[a] * y[i] * y[b])

完整代码：

/*0.089s*/

#include <cstdio>
#include <cstring>
const int MAXN = 15;

int x[MAXN], y[MAXN], d[MAXN][MAXN], path[MAXN][MAXN];

int dp(int a, int b)
{
	if (d[a][b] >= 0) return d[a][b];
	path[a][b] = a;
	if (a == b) return d[a][b] = 0;
	d[a][b] = -1u >> 1;
	int tmp;
	for (int i = a; i < b; ++i)
	{
		tmp = dp(a, i) + dp(i + 1, b) + x[a] * y[i] * y[b];
		if (tmp < d[a][b])
			d[a][b] = tmp, path[a][b] = i;
	}
	return d[a][b];
}

void print(int a, int b)
{
	if (a > b) return;
	if (a == b) printf("A%d", a + 1);
	else
	{
		printf("(");
		print(a, path[a][b]);
		printf(" x ");
		print(path[a][b] + 1, b);
		printf(")");
	}
}

int main()
{
	int n, cas = 0;
	while (scanf("%d", &n), n)
	{
		memset(d, -1, sizeof(d));
		for (int i = 0; i < n; i++)
			scanf("%d%d", &x[i], &y[i]);
		dp(0, n - 1);
		printf("Case %d: ", ++cas);
		print(0, n - 1);
		putchar(10);
	}
	return 0;
}