HDU2191 多重背包

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要点：

多重背包裸题，主要是用了两种做法，一种直接按个数全拆成01背包，复杂度是O(V*Σn[i])；一种用二进制拆成复杂度为O(V*Σlog n[i])。

复杂度是O(V*Σn[i])：

165711182016-03-16 15:17:11Accepted219131MS1416K512 BG++seasonal

#include<stdio.h>#include<string.h>#include<stdlib.h>#define max(a,b) a>b?a:bint v[150], w[150], c[150],dp[150];int n, m;int main(){int t,i,j,k;scanf("%d", &t);while (t--){memset(dp, 0, sizeof(dp));scanf("%d%d", &m, &n);for (int i = 0; i < n; i++)scanf("%d%d%d", &w[i], &v[i], &c[i]);for (i = 0; i < n; i++)for (j = 0; j < c[i]; j++)//拆成c[i]个01背包for (k = m; k >= w[i]; k--)dp[k] = max(dp[k], dp[k - w[i]] + v[i]);printf("%d\n", dp[m]);}return 0;}

复杂度为O(V*Σlog n[i])：
165707812016-03-16 14:37:38Accepted21910MS1432K985 BG++seasonal

#include<stdio.h>#include<string.h>#include<stdlib.h>#define max(a,b) a>b?a:bint v[1500], w[1500], c[1500],dp[1500];int n, m;void one_pack(int cost, int weight)//01背包{for (int j = m; j >= cost; j--)dp[j] = max(dp[j], dp[j - cost] + weight);}void complete_pack(int cost, int weight)//完全背包{for (int j = cost; j <= m; j++)dp[j] = max(dp[j], dp[j -cost] + weight);}void multiple_pack(int cost, int weight, int count){int i;if (cost*count >= m)//如果一开始就count*cost>=m，可以看成无限种也就是完全背包{complete_pack(cost, weight);}else{int k=1;while (k < count){one_pack(k*cost, k*weight);//相当于01背包count -= k;k *= 2;//利用二进制思想可以表示1-count的所有值}one_pack(count*cost, count*weight);//最后处理剩下的个数}}int main(){int t;scanf("%d", &t);while (t--){memset(dp, 0, sizeof(dp));scanf("%d%d", &m, &n);for (int i = 0; i < n; i++)scanf("%d%d%d", &w[i], &v[i], &c[i]);for (int i = 0; i < n; i++)multiple_pack(w[i], v[i], c[i]);printf("%d\n", dp[m]);}return 0;}

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