本文地址: http://blog.csdn.net/caroline_wendy
题目参考: http://blog.csdn.net/caroline_wendy/article/details/37912949
可以用动态规划(Dynamic Programming, DP)求解, 可以通过记忆化搜索推导出递推式, 可以使用三种不同的方向进行求解.
动态规划主要是状态转移, 需要理解清晰.
代码:
/* * main.cpp * * Created on: 2014.7.17 * Author: spike */ /*eclipse cdt, gcc 4.8.1*/ #include <stdio.h> #include <memory.h> #include <limits.h> #include <utility> #include <queue> #include <algorithm> using namespace std; class Program { static const int MAX_N = 100; int n=4, W=5; int w[MAX_N] = {2,1,3,2}, v[MAX_N]={3,2,4,2}; int dp[MAX_N+1][MAX_N+1]; //默认初始化为0 public: void solve() { for (int i=n-1; i>=0; i--) { for (int j=0; j<=W; ++j) { if (j<w[i]) { dp[i][j] = dp[i+1][j]; } else { dp[i][j] = max(dp[i+1][j], dp[i+1][j-w[i]] + v[i]); } } } printf("result = %d\n", dp[0][W]); } void solve1() { for (int i=0; i<n; ++i) { for (int j=0; j<=W; ++j) { if (j<w[i]) { dp[i+1][j] = dp[i][j]; } else { dp[i+1][j] = max(dp[i][j], dp[i][j-w[i]]+v[i]); } } } printf("result = %d\n", dp[n][W]); } void solve2() { for (int i=0; i<n; i++) { for (int j=0; j<=W; ++j) { dp[i+1][j] = max(dp[i+1][j], dp[i][j]); if (j+w[i]<=W) { dp[i+1][j+w[i]] = max(dp[i+1][j+w[i]], dp[i][j]+v[i]); } } } printf("result = %d\n", dp[n][W]); } }; int main(void) { Program P; P.solve2(); return 0; }
result = 7
节省空间, 可以使用1维数组的动态规划.
代码:
/* * main.cpp * * Created on: 2014.7.17 * Author: spike */ /*eclipse cdt, gcc 4.8.1*/ #include <stdio.h> #include <memory.h> #include <limits.h> #include <utility> #include <queue> #include <algorithm> using namespace std; class Program { static const int MAX_N = 100; int n=4, W=5; int w[MAX_N] = {2,1,3,2}, v[MAX_N]={3,2,4,2}; int dp[MAX_N+1]; public: void solve() { memset(dp, 0, sizeof(dp)); for (int i=0; i<n; ++i) { for (int j=W; j>=w[i]; --j) { dp[j] = max(dp[j], dp[j-w[i]]+v[i]); } } printf("result = %d\n", dp[W]); } }; int main(void) { Program P; P.solve(); return 0; }
result = 7
编程算法 - 背包问题(三种动态规划) 代码(C),布布扣,bubuko.com
原文地址:http://blog.csdn.net/caroline_wendy/article/details/37914103