UVA 348 Optimal Array Multiplication Sequence

最优矩阵链乘。

思路：根据王晓东的教材上的讲解来写的，DP入门题，最优子结构怎么来的，重叠子问题等等。个人认为思想更加比方法更加重要。

状态转移方程：

d[i][j] = 0 (i == j)

d[i][j] = min{d[i][k]+d[k][j]+R[i-1]*C[k]*C[j]} (i < j && i<=k <j)

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <string>#include <algorithm>using namespace std;const int MAXN = 15;const int INF = 0x3f3f3f3f;int R[MAXN], C[MAXN];int d[MAXN][MAXN];int path[MAXN][MAXN];int n;int times;void init(){memset(path, 0, sizeof(path));}int MatrixChain(){for(int i = 1; i <= n; i++) d[i][i] = 0;for(int r = 2; r <= n; r++) //枚举链长 {for(int i = 1; i <= n-r+1; i++) // j-i = r-1, j == n -> i <= n-r+1{int j = i+r-1; // j-i == r-1 -> j = i+r-1;d[i][j] = d[i+1][j] + R[i]*C[i]*C[j]; // k == i;path[i][j] = i;for(int k = i; k < j; k++){int t = d[i][k] + d[k+1][j] + R[i]*C[k]*C[j];if(t < d[i][j]){d[i][j] = t;path[i][j] = k;}}}}return d[1][n];}int read_case(){init();scanf("%d", &n);if(!n) return 0;for(int i = 1; i <= n; i++){scanf("%d%d", &R[i], &C[i]);}return 1;}void print_ans(int i, int j){if(i == j){printf("A%d", i);return ;}else{printf("(");print_ans(i, path[i][j]);printf(" x ");print_ans(path[i][j]+1, j);printf(")");}}void solve(){int ans = MatrixChain();printf("Case %d: ", ++times);print_ans(1, n);printf("\n");}int main(){times = 0;while(read_case()){solve();}return 0;}