HDU_1059 Dividing 【多重背包】

题目描述

解题思路

代码部分

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

const int maxn = 10;
int f[200000];
int weight[maxn], value[maxn], num[maxn];
int V, N;
/*
01背包，v为降序
f[v]:表示把前i件物品放入容量为v的背包中获得的最大收益。
f[v] = max(f[v],f[v - Weight[i]] + Value[i]);
*/
void ZeroOnePack(int nWeight,int nValue)
{
for (int v = V; v >= nWeight; v--)
{
f[v] = max(f[v],f[v - nWeight] + nValue);
}
}
/*
完全背包，v为增序。
f[v]:表示把前i件物品放入容量为v的背包中获得的最大收益。
f[v] = max(f[v],f[v - Weight[i]] + Value[i]);
*/
void CompletePack(int nWeight,int nValue)
{
for (int v = nWeight; v <= V; v++)
{
f[v] = max(f[v],f[v - nWeight] + nValue);
}
}

int MultiplePack()
{
int k = 1;
int nCount = 0;
for (int i = 1; i <= N; i++)
{
if (weight[i] * num[i] >= V)
{
//此时满足条件Weight[i] * Num[i] >= V时，
//完全背包:该类物品相当于是无限供应，直到背包放不下为止。
CompletePack(weight[i],value[i]);
}
else
{
k = 1;
nCount = num[i];
while(k <= nCount)
{
ZeroOnePack(k * weight[i],k * value[i]);
nCount -= k;
k *= 2;
}
ZeroOnePack(nCount * weight[i],nCount * value[i]);
}
}
return f[V];
}

int main()
{
int cas = 1;
while(~scanf("%d%d%d%d%d%d",&num[1],&num[2],&num[3],&num[4],&num[5],&num[6]))
{
memset(f, 0, sizeof(f));
if(num[1] == 0 && num[2] == 0&& num[3] == 0&& num[4] == 0&& num[5] == 0&& num[6] == 0)
{
break;
}
N = 6;
int sum = 0;
for(int i = 1; i <= 6; ++ i)
{
weight[i] = value[i] = i;
sum += num[i] * i;
}
printf("Collection #%d:\n", cas ++);
if(sum % 2 != 0)
{
printf("Can't be divided.\n\n");
continue;
}
bool f = false;
V = sum / 2;
if(V == MultiplePack())
{
f = true;
}
if(f)
printf("Can be divided.\n\n");
else
printf("Can't be divided.\n\n");
}
return 0;
}