挑战程序设计2 矩阵链乘

Matrix Chain Multiplication

Time Limit : 1 sec, Memory Limit : 65536 KB 
Japanese version is here

Matrix-chain Multiplication

The goal of the matrix-chain multiplication problem is to find the most efficient way to multiply given nnmatrices M1,M2,M3,...,MnM1,M2,M3,...,Mn.

Write a program which reads dimensions of MiMi, and finds the minimum number of scalar multiplications to compute the maxrix-chain multiplication M1M2...MnM1M2...Mn.

Input

In the first line, an integer nn is given. In the following nn lines, the dimension of matrix MiMi (i=1...ni=1...n) is given by two integers rr and cc which respectively represents the number of rows and columns of MiMi.

Output

Print the minimum number of scalar multiplication in a line.

Constraints

• 1≤n≤1001≤n≤100
• 1≤r,c≤1001≤r,c≤100

Sample Input 1

6
30 35
35 15
15 5
5 10
10 20
20 25

Sample Output 1

15125

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<set>
#include<string>
#include<queue>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
const int N_MAX = 200;
int n;
int dp[N_MAX][N_MAX];
int p[N_MAX];

void solve() {
    for (int i = 1; i <= n;i++) {
        dp[i][i] = 0;
     }
    for (int l = 2; l <= n;l++) {//区间长度由2开始增长
        for (int i = 1; i <= n - l + 1;i++) {
            int j = i + l - 1;//[i,j]为当前需要考虑的区间
            dp[i][j] = INF;
            for (int k = i; k <= j - 1;k++) {//!!!!!
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + p[i - 1] * p[k] * p[j]);
            }
        }
    }
    cout << dp[1][n] << endl;
}

int main() {
    while (scanf("%d",&n)!=EOF) {
        for (int i = 1; i <= n;i++) {
            scanf("%d%d",&p[i-1],&p[i]);
        }
        solve();
    }
    return 0;
}