//0-1 背包问题的动态规划解法@Java实现 分类:Source ForgeT
public class Knapsack
{
public static void knapsack(int[] v, int[] w, int c, int[][] m)
{
/** v[] w[] c 分别是价值、重量、和背包容量数组
m[i][j]表示有i~n个物品，背包容量为j的最大价值。*/
int n = v.length-1;
int jMax = Math.min(w[n]-1, c);
for(int j = 0; j <= jMax; j++)
m[n][j] = 0; //当w[n]>j 有 m[n][j]=0
//m[n][j] 表示只有n物品，背包的容量为j时的最大价值
for (int l = w[n]; l <= c; l++)
m[n][l] = v[n]; //当w[n]<=j 有m[n][j]=v[n]
//递规调用求出m[][]其它值，直到求出m[0][c]
for(int i = n-1; i >=1; i--)
{
jMax = Math.min(w[i]-1,c);
for(int k = 0; k <=jMax; k++)
m[i][k] = m[i+1][k];
for(int h = w[i]; h <= c; h++)
m[i][h] = Math.max(m[i+1][h],m[i+1][h-w[i]]+v[i]);
}
m[0][c] = m[1][c];
if(c >= w[0])
m[0][c] = Math.max(m[0][c],m[1][c-w[0]]+v[0]);
System.out.println("bestw ="+m[0][c]);
}
public static void traceback(int[][] m, int[] w, int c, int[] x)
{// 根据最优值求出最优解
int n = w.length-1;
for(int i = 0; i<n;i++)
if(m[i][c] == m[i+1][c])
x[i] = 0;
else{
x[i] = 1;
c -= w[i];
}
x[n] = (m[n][c]>0)?1:0;
}
public static void main(String[] args)
{
//测试
int[] ww = {2,2,6,5,4};
int[] vv = {6,3,5,4,6};
int[][] mm = new int[11][11];
knapsack(vv,ww,10,mm);
int[] xx =new int[ww.length];
traceback(mm,ww,10,xx);
for(int i = 0;i<xx.length;i++)
System.out.println(xx[i]);
}
}
