#include <stdio.h>#define N 1024
#define max(x, y) (x > y ? x : y)
#define min(x, y) (x < y ? x : y)void knapsack(int *w, int *v, int c, int n, int (*m)[N]);
void traceback(int *w, int (*m)[N], int *x, int n, int c);int main(int argc, const char *argv[])
{int n, c, w[N], v[N], m[N][N], x[N], i;scanf("%d%d", &n, &c);for(i = 0; i < n; i++)scanf("%d", &w[i]);for(i = 0; i < n; i++)scanf("%d", &v[i]);knapsack(w, v, c, n-1, m);      //这里传的是数组的最大下标traceback(w, m, x, n-1, c);     //求出是否装载的序列 x[i]printf("Max v: %d\n", m[0][c]);for(i = 0; i < n; i++)printf("x[%d] = %d\t", i, x[i]);printf("\n");return 0;
}void knapsack(int *w, int *v, int c, int n, int (*m)[N]){int j, jMax, i;jMax = min(w[n] - 1, c);for(j = 0; j <= jMax; j++)      //求m[n][]m[n][j] = 0;for(j = c; j > jMax; j--)m[n][j] = v[n];for(i = n-1; i > 0; i--){       //依次求m[n-1][] - m[1]jMax = min(w[i]-1, c);for(j = 0; j <= jMax; j++){m[i][j] = m[i+1][j];}for(j = c; j > jMax; j--){m[i][j] = max(m[i+1][j], m[i+1][j-w[i]] + v[i]);}}m[0][c] = m[1][c];              //求m[0][]if(w[0] < c){m[0][c] = max(m[1][c], m[1][c-w[0]] + v[0]);}
}void traceback(int *w, int (*m)[N], int *x, int n, int c){int i;for(i = 0; i < n-1; i++){if(m[i][c] == m[i+1][c])    //根据m数组 判断是否装进去x[i] = 0;else{x[i] = 1;c -= w[i];}}x[n] = (m[n][c] > 0) ? 1 : 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <string.h>#define N 5
#define C 10int w[] = {2, 2, 6, 5,4};           //使用递归 避免传递太多参数，不清晰
int v[] = {6, 3, 5, 4,6};           //因此设置成全局变量
int m[N][C], x[N];int knapsack(int i, int j);
void traceback();int main(int argc, const char *argv[])
{int i;knapsack(0, C);traceback();printf("Max :%d\n", m[0][10]);for(i = 0; i < N; i++)printf("x[%d] = %d\t", i, x[i]);printf("\n");return 0;
}int knapsack(int i, int j){if(i == N-1){m[i][j] = (j > w[i] ? v[i] : 0);return  m[i][j];}int ret1, ret2;if(j < w[i]){m[i][j] = knapsack(i+1, j);}else{ret1 = knapsack(i+1, j);ret2 = knapsack(i+1, j-w[i]) + v[i];m[i][j] =  ret1 > ret2 ? ret1 : ret2;}return m[i][j];}void traceback(){int i, n = N-1, c = C;for(i = 0; i < n; i++){if(m[i][c] == m[i+1][c])    //根据m数组 判断是否装进去x[i] = 0;else{x[i] = 1;c -= w[i];}}x[n] = (m[n][c] > 0) ? 1 : 0;
}