1、因此0/1背包问题是一个特殊的整数规划问题。【算法设计】设0/1背包问题的最优值为m( i, j ),即背包容量是j,可选择物品为i,i+1,n时0/1背包问题的最优值。由0/1背包问题的最优子结构性质,可以建立计算m( i, j )的递归式如下: maxm( i+1, j ), m( i+1, j-)+ m( i, j )=m(i+1,j) m(n,j)=0 【算法实现】#include #include iomanip.hint min(int w, int c) int temp;if (w void knapsack(int v, int w, int* m, int c, int n
2、)/求最优值 int jmax = min(wn-1, c);for (int j = 0; j = jmax; j+) mnj = 0;for (int jj = wn; jj 1; i-)/递归部分 jmax = min(wi-1, c);for(int j = 0;mij = mi+1j;for(int jj = wi;mijj = max(mi+1jj, mi+1jj-wi+vi); m1c = m2c;if(c = w1) m1c = max(m1c, m2c-w1+v1);cout endl 最优值: m1c endl;coutendl;&int traceback(int x,
3、int w, int* m, int c, int n) /回代,求最优解 out 得到的一组最优解如下:for(int i = 1; i n; i+) if(mic = mi+1c)xi = 0;else xi = 1;c -= wi; xn = (mnc) ? 1:0;for(int y = 1; y = n; y+) cout xy n ;请输入背包的承重: c;int *v = new intn+1;请输入每个物品的价值 (vi):cin vi;int *w = new intn+1;请输入每个物品的重量 (wi):for(int j = 1; wj;int *x = new intn+1;m = new int* n+1;/动态的分配二维数组 for(int p = 0; p n+1; p+) mp = new intc+1;knapsack (v, w, m, c, n);traceback(x, w, m, c, n);【运行结果】