多重背包（二进制模板+普通解法）
题目：http://acm.hdu.edu.cn/showproblem.php?pid=2191
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 6 using namespace std; 7 const int MAX = 100 + 10; 8 int V; 9 int f[MAX],Price[MAX],Weight[MAX],Amount[MAX]; 10 void OneZeorPage(int cost, int weight) 11 { 12 for(int i = V; i >= cost; i--) 13 f[i] = max(f[i], f[i - cost] + weight); 14 } 15 void CompletePage(int cost, int weight) 16 { 17 for(int i = cost; i <= V; i++) 18 f[i] = max(f[i], f[i - cost] + weight); 19 20 } 21 void multiplePage(int cost,int weight,int amount) 22 { 23 if(amount * cost >= V) 24 { 25 CompletePage(cost,weight); 26 return; 27 } 28 int k = 1; 29 while(k < amount) 30 { 31 OneZeorPage(k * cost, k * weight); 32 amount -= k; 33 k <<= 1; 34 } 35 if(amount > 0) 36 OneZeorPage(amount * cost, amount * weight); 37 return; 38 } 39 int main() 40 { 41 int n,m,t; 42 scanf("%d", &t); 43 while(t--) 44 { 45 scanf("%d%d", &n,&m); 46 for(int i = 1; i <= m; i++) 47 scanf("%d%d%d", &Price[i],&Weight[i],&Amount[i]); 48 memset(f,0,sizeof(f)); 49 V = n; 50 for(int i = 1; i <= m; i++) 51 multiplePage(Price[i],Weight[i],Amount[i]); 52 printf("%d\n",f[n]); 53 } 54 return 0; 55 }
那为什么会有完全背包和01 背包的不同使用加判断呢？原因也很简单啊，当数据很大，大于背包的容纳量时，我们就是在这个物品中取上几件就是了，取得量时不知道的，也就理解为无限的啦，这就是完全背包啦，反而小于容纳量的就是转化为01背包来处理就是了，可以大量的省时间。
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 6 using namespace std; 7 const int MAX = 100 + 10; 8 int V; 9 int f[MAX],Price[MAX],Weight[MAX],Amount[MAX]; 10 int main() 11 { 12 int n,m,t; 13 scanf("%d", &t); 14 while(t--) 15 { 16 scanf("%d%d", &n,&m); 17 for(int i = 1; i <= m; i++) 18 scanf("%d%d%d", &Price[i],&Weight[i],&Amount[i]); 19 memset(f,0,sizeof(f)); 20 V = n; 21 for(int i = 1; i <= m; i++) 22 { 23 for(int k = 1; k <= Amount[i]; k++) 24 { 25 for(int j = V; j >= Price[i]; j--) 26 { 27 f[j] = max(f[j], f[j - Price[i]] + Weight[i]); 28 } 29 } 30 } 31 printf("%d\n",f[n]); 32 } 33 return 0; 34 }
