348 - Optimal Array Multiplication Sequence
Time limit: 3.000 seconds
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;
}

本文介绍了一种使用记忆化搜索解决最优矩阵链乘法问题的方法。通过递归地寻找最佳分割点来最小化所需的乘法次数,并记录下计算结果避免重复计算。最终,不仅能找出最少的乘法次数,还能打印出最优的矩阵乘法顺序。
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