UVa 348 - Optimal Array Multiplication Sequence

矩阵链乘的问题，好像很多算法书上讲DP都是开始就将矩阵链乘。这个题不是要求最小的运算量，而是求最优的方案是什么，难就难在了递归的输出，如何记录路径，如何输出。

我也是参看了别人的代码。递归输出。不过这个转移方程还是蛮好找的。dp[i][j] = min ( dp[i][k], dp[k + 1][j], p[i - 1]p[k]p[j] ).

// File Name: UVa348.cpp// Author: Toy// Created Time: 2013年04月15日 星期一 19时03分31秒#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <cctype>#include <cmath>#include <string>#include <algorithm>#include <cstdlib>#include <iomanip>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <utility>#include <bitset>#define L(x) x << 1#define R(x) x << 1 | 1using namespace std;int num, path[15][15], p[15], dp[15][15], Case, tmp;void output ( int i, int j ) {    if ( i == j ) printf ( "A%d", i );    else {printf ( "(" );output ( i, path[i][j] );printf ( " x " );output ( path[i][j] + 1, j );printf ( ")" );    }}void DP ( ) {    for ( int d = 2; d <= num; ++d ) {for ( int i = 1, j = d; j <= num; ++i, ++j ) {    dp[i][j] = 10000000000;    for ( int k = i; k < j; ++k ) {tmp = dp[i][k] + dp[k + 1][j] + p[i - 1] * p[j] * p[k];if ( dp[i][j] > tmp ) dp[i][j] = tmp, path[i][j] = k;    }}    }}int main ( ) {    Case = 0;    while ( scanf ( "%d", &num ) != EOF && num ) {for ( int i = 1; i <= num; ++i )     scanf ( "%d%d", &p[i - 1], &p[i] );memset ( dp, 0, sizeof ( dp ) );memset ( path, 0, sizeof ( path ) );DP ( );printf ( "Case %d: ", ++Case );output ( 1, num );printf ( "\n" );    }    return 0;}