最少攔截系統
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44681 Accepted Submission(s): 17510
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.
// 可以有兩種方法: 1 ) 用貪心算法 ; 2 ) LIS ;
//貪心代碼如下:
/*
如果輸入爲 5 100 97 98 96 97
*/
#include <stdio.h>
const int maxn = 300010;
int n;
int a[maxn];
int fun(int n)
{
for (int i = 0 ; i < n ; i++)
if (a[i] != -1)
return 1;
return 0;
}
int main()
{
while(~scanf ("%d",&n))
{
for (int i = 0 ; i < n ; i++)
scanf ("%d",&a[i]);
int ans = 0;
int k;
int temp;
int fun(int n);
while (fun(n))
{
ans++;
temp = 0;
for (int i = 0 ; i < n ; i++)
{
if (a[i] != -1)
{
temp = a[i];
a[i] = -1;
k = i;
break;
}
}
for (int i = k + 1 ; i < n ; i++)
{
if (a[i] != -1 && a[i] <= temp)
{
temp = a[i];
a[i] = -1;
}
}
}
printf ("%d\n",ans);
}
return 0;
}
//LIS代碼如下:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 30010;
int dp[maxn], a[maxn];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;++i)
scanf("%d",&a[i]);
int ans=0;
for(int i=0;i<n;++i)
{
dp[i]=1;
for(int j=0;j<i;++j)
{
if(a[j]<a[i])
dp[i] = max(dp[i], dp[j]+1);
}
ans=max(dp[i],ans);
}
printf("%d\n",ans);
}
return 0;
}
//另一種LIS代碼如下:
#include <cstdio>
#include <algorithm>
#define INF 0x3f3f3f
using namespace std;
int dp[40010],a[40010];
int main()
{
int i,j;
int t,n;
while(~scanf ("%d",&n))
{
for(i=0;i<n;++i)
{
scanf("%d",&a[i]);
dp[i]=INF;
}
for(i=0;i<n;++i)
*lower_bound(dp,dp+n,a[i])=a[i];
printf("%d\n",lower_bound(dp,dp+n,INF)-dp);
}
return 0;
}