ZOJ 1602 Multiplication Puzzle(矩阵连乘)

来源：互联网发布：vb计算器代码编辑：程序博客网时间：2024/05/17 01:44

题意:给出一排卡,每次能拿走除了第一张和最后一张之外的任何卡,拿走卡i之后要加上卡i和卡i-1和卡i+1的乘积,直到剩下第一张和最后一张为止,求最少的乘积和.

设dp[i][j]为拿走i..j之间卡的最少乘积和,那么答案就是dp[1][N]

dp[i][j] = min{dp[i][k[ + dp[k][j] + C[i] * C[k] * C[j]] | i< k < j}

base cases: dp[i][i] = 0, dp[i][i + 1] = 0.

#include <cstdio>#include <memory.h>#include <algorithm>using namespace std;const int MAX = 105;int main(int argc, char const *argv[]){int dp[MAX][MAX];int cards[MAX];int N;while(scanf("%d", &N) == 1){memset(dp, 0x20, sizeof(dp));for(int i = 1; i <= N; ++i){scanf("%d", &cards[i]);dp[i][i] = dp[i][i + 1] = 0;}for(int p = 2; p < N; ++p){for(int i = 1; i <= N && i + p <= N; ++i){int j = i + p;for(int k = i; k <= j; ++k){dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + cards[i] * cards[k] * cards[j]);}}}printf("%d\n", dp[1][N]);}return 0;}

0 0