编程算法 - 多重部分和问题 代码(C)

多重部分和问题 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

题目: 有n种不同大小的数字a, 每种各m个. 推断能否够从这些数字之中选出若干使它们的和恰好为K.

使用动态规划求解(DP),

方法1: dp[i+1][j] = 用前n种数字能否加和成j, 时间复杂度O(nKm), 不是最优.

方法2: dp[i+1][j] = 用前i种数加和得到j时, 第i种数最多能剩余多少个. 时间复杂度O(nK).

比如: n=3, a={3,5,8}, m={3,2,2}, K=17时.

i\j	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17
起始	0	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1
0(3,3)	3	-1	-1	2	-1	-1	1	-1	-1	0	-1	-1	-1	-1	-1	-1	-1	-1
1(5,2)	2	-1	-1	2	-1	1	2	-1	1	2	0	-1	-1	0	1	-1	-1	-1
2(8,2)	2	-1	-1	2	-1	2	2	-1	2	2	2	1	-1	1	1	-1	1	1

代码:

/*
 * main.cpp
 *
 *  Created on: 2014.7.20
 *      Author: spike
 */

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <memory.h>

class Program {
	static const int MAX_N = 100;
	int n = 3;
	int K = 17;
	int a[MAX_N] = {3,5,8};
	int m[MAX_N] = {3,2,2};
	bool dp[MAX_N+1][MAX_N+1];
public:
	void solve() {
		dp[0][0] = true;
		for (int i=0; i<n; ++i) {
			for (int j=0; j<=K; ++j) {
				for (int k=0; k<=m[i]&&k*a[i]<=j; ++k) {
					dp [i+1][j] |= dp[i][j-k*a[i]]; //或运算
				}
			}
		}
		if (dp[n][K]) printf("result = Yes\n");
		else printf("result = No\n");
	}
};

class Program2 {
	static const int MAX_N = 100;
	static const int MAX_K = 20;
	int n = 3;
	int K = 17;
	int a[MAX_N] = {3,5,8};
	int m[MAX_N] = {3,2,2};
	int dp[MAX_K+1];
public:
	void solve() {
		memset(dp, -1, sizeof(dp));
		dp[0] = 0;
		for (int i=0; i<n; ++i) {
			for (int j=0; j<=K; ++j) {
				if (dp[j] >= 0) {
					dp[j] = m[i];
				} else if (j < a[i] || dp[j-a[i]]<=0){
					dp[j] = -1;
				} else {
					dp[j] = dp[j-a[i]]-1;
				}
			}
		}
		if (dp[K]>=0) printf("result = Yes\n");
		else printf("result = No\n");
	}
};

int main(void)
{
	Program2 iP;
	iP.solve();

	return 0;
}

输出:

result = Yes