POJ1651Multiplication Puzzle（区间DP）
比较好做的区间DP
状态转移方程：DP[i][j] 表示区间[i,j]最小的乘积和。
DP[i][j] = MIN{DP[i][k-1]+DP[k+1][j] + a[k]*a[i-1]*a[j+1] | i<k<j};
最后结果就是DP[2][N-1]
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define eps 1e-15 16 #define MAXN 105 17 #define INF 1000000007 18 #define MAX(a,b) (a > b ? a : b) 19 #define MIN(a,b) (a < b ? a : b) 20 #define mem(a) memset(a,0,sizeof(a)) 21 22 int a[105],DP[105][105],N; 23 24 int main() 25 { 26 while(~scanf("%d", &N)) 27 { 28 for(int i=0;i<=N;i++) 29 { 30 for(int j=0;j<=N;j++) DP[i][j] = INF; 31 } 32 for(int i=1;i<=N;i++) 33 { 34 scanf("%d", &a[i]); 35 } 36 for(int i=2;i<N;i++) 37 { 38 for(int j=i;j>=2;j--) 39 { 40 if(i == j) DP[j][i] = a[i] * a[i-1] * a[i+1]; 41 else 42 { 43 int x = DP[j+1][i]+a[j]*a[j-1]*a[i+1]; 44 int y = DP[j][i-1]+a[i]*a[j-1]*a[i+1]; 45 DP[j][i] = MIN(x, y); 46 if(i-j>1)for(int k = j+1;k<i;k++) 47 { 48 x = DP[j][k-1] + DP[k+1][i] + a[k]*a[j-1]*a[i+1]; 49 DP[j][i] = MIN(DP[j][i], x); 50 } 51 } 52 } 53 } 54 printf("%d\n", DP[2][N-1]); 55 } 56 return 0; 57 }
