您的位置:首页 > 其它

背包问题模板,POJ(1014)

2016-05-26 20:31 309 查看
题目链接:http://poj.org/problem?id=1014

背包问题太经典了,之前的一篇博客已经讲了背包问题的原理。

这一个题目是多重背包,但是之前的枚举是超时的,这里采用二进制优化。

这是所有01背包,完全背包,多重背包的模板哦!

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int sum;
int num[7], dp[60000 + 60];

void ZeroOnePack(int cost, int weight, int V)
{
for (int i = V; i >= cost; i--)
{
dp[i] = max(dp[i], dp[i - cost] + weight);
}
}

void CompletePack(int cost, int weight, int V)
{
for (int i = cost; i <= V; i++)
{
dp[i] = max(dp[i], dp[i - cost] + weight);
}
}

void MultiPack(int cost, int weight, int V, int amount)
{
if (cost * amount >= V)
{
CompletePack(cost, weight, V);
return;
}
int k = 1;
while (k < amount)
{
ZeroOnePack(cost * k, weight * k, V);
amount -= k;
k *=2;
}
ZeroOnePack(cost * amount, weight * amount, V);
}

int main()
{
int t = 1;
while (~scanf("%d", &num[1]))
{
sum = num[1];
for (int i = 2; i <= 6; i++)
{
scanf("%d", &num[i]);
sum += num[i] * i;
}
if (num[1] + num[2] + num[3] + num[4] + num[5] + num[6] == 0) break;
printf("Collection #%d:\n", t++);
if (sum % 2)
{
puts("Can't be divided.\n");
continue;
}
sum >>= 1;
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= 6; i++)
{
MultiPack(i, i, sum, num[i]);
}
if (dp[sum] != sum)
{
puts("Can't be divided.\n");
}
else
puts("Can be divided.\n");
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: