问题:
设N个矩阵序列,其中第i个矩阵式p[i - 1] * p[i]阶矩阵,给定矩阵的向量P,求一种乘法次序,使得基本运算总次数最小
解析:
设A[i][j] 为 ∏jk = ia[k]
设cnt[i][j]为A[i][j]的最少运算次数
cnt[i][j] = min(cnt[i][k] + cnt[k + 1][j] + p[i - 1] * p[k] * p[j])
设计(核心代码):
1 for (int len = 2; len <= n; ++len) 2 { 3 for (int i = 1; i + len - 1 <= n; ++i) 4 { 5 int j = i + len - 1; 6 cnt[i][j] = inf; 7 s[i][j] = 0; 8 for (int k = i; k < j; ++k) 9 { 10 int tmp = cnt[i][k] + cnt[k + 1][j] + p[i - 1] * p[k] * p[j]; 11 if (tmp < cnt[i][j]) 12 { 13 cnt[i][j] = tmp; 14 s[i][j] = k; 15 } 16 } 17 } 18 }
分析:
O(n3)
源码:
https://github.com/BambooCertain/Algorithm.git
1 #include<cstdio> 2 #include<string.h> 3 #include<algorithm> 4 #include<cmath> 5 #include<iostream> 6 #include<vector> 7 #include<queue> 8 #include<set> 9 #include<map> 10 #include<cctype> 11 #define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) 12 #define mem(a,x) memset(a,x,sizeof(a)) 13 #define lson rt<<1,l,mid 14 #define rson rt<<1|1,mid + 1,r 15 #define P pair<int,int> 16 #define ull unsigned long long 17 using namespace std; 18 typedef long long ll; 19 const int maxn = 2e5 + 10; 20 const ll mod = 998244353; 21 const int inf = 0x3f3f3f3f; 22 const long long INF = 0x3f3f3f3f3f3f3f3f; 23 const double eps = 1e-7; 24 25 inline ll read() 26 { 27 ll X = 0, w = 0; char ch = 0; 28 while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); } 29 while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar(); 30 return w ? -X : X; 31 } 32 33 int cnt[1000][1000] , s[1000][1000]; 34 int p[1000]; 35 int n; 36 int main() 37 { 38 n = read(); 39 for (int i = 0; i <= n; ++i) 40 { 41 p[i] = read(); cnt[i][i] = 0; 42 } 43 for (int len = 2; len <= n; ++len) 44 { 45 for (int i = 1; i + len - 1 <= n; ++i) 46 { 47 int j = i + len - 1; 48 cnt[i][j] = inf; 49 s[i][j] = 0; 50 for (int k = i; k < j; ++k) 51 { 52 int tmp = cnt[i][k] + cnt[k + 1][j] + p[i - 1] * p[k] * p[j]; 53 if (tmp < cnt[i][j]) 54 { 55 cnt[i][j] = tmp; 56 s[i][j] = k; 57 } 58 } 59 } 60 } 61 cout << cnt[1][n] << endl; 62 return 0; 63 }