ZOJ 1602 Multiplication Puzzle

区间dp

F(i, j) = min(F(i, k) + a[i]*a[k]*a[j] + F(k, j))

#include <cstdio>
#include <cstring>
using namespace std;
#define N 103
typedef long long LL;
const long long inf = 1<<29;
LL f[N][N], a[N];
int n;
LL dp(int l, int r) {
    if (f[l][r] != inf) return f[l][r];
    if (l + 2 > r) return f[l][r] = 0;
    LL min = inf, tmp;
    for (int k=l+1; k<r; k++) {
        tmp = dp(l, k) + dp(k, r) + a[l]*a[k]*a[r];
        if (tmp < min) min = tmp;
    }
    return f[l][r] = min;
}
int main() {
    while (scanf("%d", &n) == 1) {
        for (int i=1; i<=n; i++) scanf("%lld", &a[i]);
        for (int i=1; i<=n; i++)
            for (int j=1; j<=n; j++) f[i][j] = inf;
        printf("%lld\n", dp(1, n));
    }
    return 0;
}