題目大意:有N種東西,現已知每樣東西的價值和數量,將N種東西分成兩堆,且保證第一堆的價值不少於第二堆的前提下,使兩堆的價值儘可能相等
解題思路:
考慮,由於第二堆的價值小於等於第一堆,也就是說第二堆的價值的最大值不能超過總價值的一般,把這個看成揹包的容量,可以得到如下的狀態轉移方程
value[j]=max{value[j],value[j-facility[i].v]+facility[i].v}
代碼如下:
# include <iostream>
# include <algorithm>
using namespace std;
struct node
{
int v,m;
}facility[10002];
int value[300002];
int main()
{
freopen("input.txt","r",stdin);
int n;
while((scanf("%d",&n)!=EOF) && n>=0)
{
int i,j,k;
int sum=0;
for(i=0;i<n;i++)
{
scanf("%d %d",&facility[i].v,&facility[i].m);
sum=sum+facility[i].v*facility[i].m;
}
int half=sum/2;
memset(value,0,sizeof(value));
for(i=0;i<n;i++)
{
for(j=1;j<=facility[i].m;j++)
{
for(k=half;k>=facility[i].v;k--)
{
if(value[k]<(value[k-facility[i].v]+facility[i].v))
{
value[k]=value[k-facility[i].v]+facility[i].v;
}
}
}
}
printf("%d %d\n",sum-value[half],value[half]);
}
return 0;
}