UVALive5583 UVA562 Dividing coins

Regionals 1996 >> Europe-northwestern

Problem Link: UVALive5583 UVA562 dividing coins.

Test Instructions Summary: Enter the number of test cases N, each test case includes the number of gold coins m and m positive integers are the value of gold coins. Divide these coins into two piles to minimize the difference and to find the smallest difference.

Problem analysis: You can solve this problem with the dynamic programming method of 0/1 knapsack . Set the backpack's capacity to half the sum of all gold coins, and put the pile of gold coins in the backpack as much as possible. Then calculate the difference.

0/1 Backpack Dynamic Programming method of the recursive type F (j,x), wherein the j=0,1,2,..., n-1 for items (total n items), X is the load capacity of the backpack. f (j,x) means: When the backpack load is X, the optional item is the optimal value of 0,1,2,..., J. The following are the specific formulas:

F ( -1,x) =-∞(x<0), F ( -1,x) =0 (x>=0)

F (j,x) =max{f (j-1,x), F (j-1 (X-WJ) +PJ} 0<=j<n,wj for Article J Item weight, PJ for article J article proceeds.

The program uses a two-dimensional array dp[][] to hold the value of the recursive function, compared to waste storage space, it should be used to construct a sparse matrix to store more reasonable.

In this procedure, Wj=coin[j] is the value of the currency, Pj=coin[j] that is the currency, a special case. The program logic is actually applied to the 0/1 knapsack recursive function.

The C language Program of AC is as follows:

/* UVALive5583 UVA562 dividing coins */#include <stdio.h> #include <memory.h> #define MAX (x, Y) (((x) > (y))? x):(y)) #define MAXN 100int DP[MAXN+1][MAXN * 500];int main (void) {    int t, M, coin[maxn+1], sum, sum2, I, J;    scanf ("%d", &t);    while (t--) {        scanf ("%d", &m);        memset (coin, 0, sizeof (coin));        memset (DP, 0, sizeof (DP));        sum = 0;        for (I=1; i<=m; i++) {            scanf ("%d", &coin[i]);            Sum + = Coin[i];        }        sum2 = SUM/2;        for (I=1; i<=m; i++)            for (j=0; j<=sum2; j + +)                if (J >= Coin[i])                    dp[i][j] = MAX (dp[i-1][j), (dp[i-1 ][j-coin[i]] (coin[i]));                else                    dp[i][j] = dp[i-1][j];        printf ("%d\n", sum-dp[m][sum2]-dp[m][sum2]);    }    return 0;}

UVALive5583 UVA562 Dividing coins