最少攔截系統
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34275 Accepted Submission(s): 13487
怎麼辦呢?多搞幾套系統唄!你說說倒蠻容易,成本呢?成本是個大問題啊.所以俺就到這裏來求救了,請幫助計算一下最少需要多少套攔截系統.
最長單調遞增子序列模板題,分別用O(N*N)和O(N*logN)兩種方法。
法一(O(N*N)):dp[i]是爲以a[i]爲結尾的最長遞增子序列的長度,可得到遞推關係:
dp[i]=max(dp[i],dp[j]+1);
AC代碼:
#include<cstdio>
#include<cstring>
#define MAXN 40005
int a[MAXN],b[MAXN];
int search(int num,int low,int high)
{
int mid;
while(low<=high)
{
mid=(low+high)/2;
if(num>=b[mid])
low=mid+1;
else
high=mid-1;
}
return low;
}
int main()
{
//freopen("D://in.txt", "r", stdin);
int n,len,pos;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
b[1]=a[1];
len=1;
for(int i=2;i<=n;i++)
{
if(a[i]>=b[len])
b[++len]=a[i];
else
{
pos=search(a[i],1,len);
b[pos]=a[i];
}
}
printf("%d\n",len);
}
return 0;
}