本文地址: http://blog.csdn.net/caroline_wendy
题目参考: http://blog.csdn.net/caroline_wendy/article/details/37912949
使用记忆化搜索, 需要存储每组的值, 下次不需要进行继续迭代, 可以降低至时间复杂度O(nW).
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];
int rec (int i, int j) {
if (dp[i][j] >= 0) return dp[i][j];
int res;
if (i==n) {
res = 0;
} else if (j<w[i]) {
res = rec(i+1, j);
} else {
res = max(rec(i+1,j), rec(i+1, j-w[i])+v[i] );
}
return dp[i][j] = res;
}
public:
void solve() {
memset(dp, -1, sizeof(dp));
printf("result = %d\n", rec(0, W));
}
};
int main(void)
{
Program P;
P.solve();
return 0;
}
result = 7
编程算法 - 背包问题(记忆化搜索) 代码(C),布布扣,bubuko.com
原文地址:http://blog.csdn.net/caroline_wendy/article/details/37913341