Programming algorithms-Multiple parts and problem codes (C)

multiple parts and problem codes (C)

This address: Http://blog.csdn.net/caroline_wendy

Title: There are n different sizes of numbers a, each of each m. The inference can be enough to select some of these numbers to make their and exactly K.

Using dynamic Solver (DP),

Method 1: dp[i+1][j] = The number of the first n can be added to J, time Complexity O (nKm), is not optimal.

Method 2: dp[i+1][j] = When the number of first I is added and the J is obtained, the number of the first I can be the maximum number of remaining. time Complexity O (NK).

For example: N=3, a={3,5,8}, m={3,2,2}, k=17.

I\j	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17
Starting	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

Code:

/* * 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];p ublic: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]];//or Operation}}}if (Dp[n][k]) printf ("Re Sult = 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];p ublic:void Solve () {memset (DP,-1, sizeof (DP));DP [0] = 0;for (int i=0; i<n; ++i) {fo R (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;} 

Output:

result = Yes

Programming algorithms-Multiple parts and problem codes (C)