Multiplication Puzzle POJ - 1651

版权声明： https://blog.csdn.net/weixin_40959045/article/details/88382132

1. 显然1,N是要会留下的
2. dp[l][r] 代表消去l,r区间内的数所能获取的最小值
3. 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;
}