版权声明: https://blog.csdn.net/weixin_40959045/article/details/88382132
- 显然1,N是要会留下的
- dp[l][r] 代表消去l,r区间内的数所能获取的最小值
- 枚举区间内每一个点
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define fst first
#define sec second
#define sci(num) scanf("%d",&num)
#define scl(num) scanf("%lld",&num)
#define mem(a,b) memset(a,b,sizeof a)
#define cpy(a,b) memcopy(a,b,sizeof b)
typedef long long LL;
typedef pair<int,int> P;
const int MAX_N = 110;
const int MAX_M = 10000;
int dp[MAX_N][MAX_N];
int nums[MAX_N];
int main() {
int N;
sci(N);
for (int i = 1;i <= N;i++)
sci(nums[i]);
mem(dp,0x3f3f3f3f);
for (int i = 1;i <= N;i++)
dp[i][i-1] = 0;
for (int i = 2;i < N;i++)
dp[i][i] = nums[i - 1] * nums[i] * nums[i + 1];
for (int len = 1;len < N;len++) {
for (int i = 1;i + len <= N;i++) {
for (int j = i;j <= i + len;j++) {
if (j == 1 || j == N) continue;
dp[i][i + len] = min(dp[i][i + len],
dp[i][j-1]+dp[j+1][i+len]+nums[i-1] * nums[j] * nums[i + len +1]);
}
}
}
printf("%d\n",dp[2][N-1]);
return 0;
}