Description
In the army, a platoon is composed by n soldiers. During the morning inspection, the soldiers are aligned in a straight line in front of the captain. The captain is not satisfied with the way his soldiers are aligned; it is true that the soldiers are aligned in order by their code number: 1 , 2 , 3 , . . . , n , but they are not aligned by their height. The captain asks some soldiers to get out of the line, as the soldiers that remain in the line, without changing their places, but getting closer, to form a new line, where each soldier can see by looking lengthwise the line at least one of the line's extremity (left or right). A soldier see an extremity if there isn't any soldiers with a higher or equal height than his height between him and that extremity.
Write a program that, knowing the height of each soldier, determines the minimum number of soldiers which have to get out of line.
Write a program that, knowing the height of each soldier, determines the minimum number of soldiers which have to get out of line.
Input
On the first line of the input is written the number of the soldiers n. On the second line is written a series of n floating numbers with at most 5 digits precision and separated by a space character. The k-th number from this line represents the height of the soldier who has the code k (1 <= k <= n).
There are some restrictions:
• 2 <= n <= 1000
• the height are floating numbers from the interval [0.5, 2.5]
There are some restrictions:
• 2 <= n <= 1000
• the height are floating numbers from the interval [0.5, 2.5]
Output
The only line of output will contain the number of the soldiers who have to get out of the line.
Sample Input
8 1.86 1.86 1.30621 2 1.4 1 1.97 2.2
Sample Output
4
题解：可以直接将这个问题轻松转化为01背包，但复杂度为V*∑n[i]比较高，所以需要优化：
解题过程：对于每个物品，有三种情况：
①：单件该物品体积已经超过背包容量，不可能装的进
②：无法将该物品全部装入背包，直接将它当做完全背包
③：可以将该物品全部装入背包，这种情况比较复杂，可以将这些物品拆成多个单件物品，但复杂度很高（V*∑n[i]），所以需要二进制优化，
假设有30个该物品，每件物品体积为10，将这30个物品拆成5个单件物品即可，它们的体积分别为10、20、40、80、150，这样对于任意的m(m<=30)
件物品都可以用这5个物品表示出来
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[100005], V;
void Alone(int use)
{
int v;
for(v=V;v>=use;v--)
dp[v] = max(dp[v], dp[v-use]+use);
}
void Every(int use)
{
int v;
for(v=use;v<=V;v++)
dp[v] = max(dp[v], dp[v-use]+use);
}
void Mulitp(int n, int use)
{
int k;
if(use>V)
return;
if(use*n>=V)
Every(use);
else
{
k = 1;
while(k<n)
{
Alone(k*use);
n -= k;
k *= 2;
}
Alone(n*use);
}
}
int main(void)
{
int i, n, s[15], val[15];
while(scanf("%d", &V)!=EOF)
{
memset(dp, 0, sizeof(dp));
scanf("%d", &n);
for(i=1;i<=n;i++)
scanf("%d%d", &s[i], &val[i]);
for(i=1;i<=n;i++)
Mulitp(s[i], val[i]);
printf("%d\n", dp[V]);
}
}
