矩阵连乘积 ZOJ 1276 Optimal Array Multiplication Sequence

题目传送门

/*
    题意：加上适当的括号，改变计算顺序使得总的计算次数最少
    矩阵连乘积问题，DP解决：状态转移方程：
    dp[i][j] = min (dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j])    (i<=k<j)
    s[i][j] 记录断开的地方(即加括号的位置)，回溯法输出结果
*/
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <iostream>
using namespace std;

const int MAXN = 1e2 + 10;
const int INF = 0x3f3f3f3f;
int dp[MAXN][MAXN];
int s[MAXN][MAXN];
int p[MAXN];
int n;

void print(int i, int j)
{
    if (i == j)    printf ("A%d", i);
    else
    {
        printf ("(");
        print (i, s[i][j]);
        printf (" x ");
        print (s[i][j] + 1, j);
        printf (")");
    }
}

void work(void)
{
    for (int i=1; i<=n; ++i)    dp[i][i] = 0;
    for (int l=2; l<=n; ++l)
    {
        for (int i=1; i<=n-l+1; ++i)
        {
            int j = i + l - 1;
            dp[i][j] = INF;
            for (int k=i; k<=j-1; ++k)
            {
                int tmp = dp[i][k] + dp[k+1][j] + p[i-1] * p[k] * p[j];
                if (tmp < dp[i][j])
                {
                    dp[i][j] = tmp;    s[i][j] = k;
                }
            }
        }
    }

    print (1, n);    puts ("");
}

int main(void)        //ZOJ 1276 Optimal Array Multiplication Sequence
{
    //freopen ("ZOJ_1276.in", "r", stdin);

    int cas = 0;
    while (scanf ("%d", &n) == 1)
    {
        if (n == 0)    break;
        for (int i=1; i<=n; ++i)    scanf ("%d%d", &p[i-1], &p[i]);

        printf ("Case %d: ", ++cas);
        work ();
    }

    return 0;
}

/*
Case 1: (A1 x (A2 x A3))
Case 2: ((A1 x A2) x A3)
Case 3: ((A1 x (A2 x A3)) x ((A4 x A5) x A6))
*/