POJ2533:Longest Ordered Subsequence(LIS)
Description
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
Input
Output
Sample Input
7
1 7 3 5 9 4 8
Sample Output
4
题意:输出最长递增子序列的长度
思路:直接裸LIS,第一次使用,使用两种方法
第一种:复杂度n^2
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std; int a[1005],dp[1005],n; int LIS()
{
int i,j,ans,m;
dp[1] = 1;
ans = 1;
for(i = 2;i<=n;i++)
{
m = 0;
for(j = 1;j<i;j++)
{
if(dp[j]>m && a[j]<a[i])
m = dp[j];
}
dp[i] = m+1;
if(dp[i]>ans)
ans = dp[i];
}
return ans;
} int main()
{
int i;
while(~scanf("%d",&n))
{
for(i = 1;i<=n;i++)
scanf("%d",&a[i]);
printf("%d\n",LIS()); } return 0;
}
第二种:nlogn
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std; int a[1005],dp[1005],c[1005],n; int bin(int size,int k)
{
int l = 1,r = size;
while(l<=r)
{
int mid = (l+r)/2;
if(k>c[mid] && k<=c[mid+1])
return mid+1;
else if(k<c[mid])
r = mid-1;
else
l = mid+1;
}
} int LIS()
{
int i,j,ans=1;
c[1] = a[1];
dp[1] = 1;
for(i = 2; i<=n; i++)
{
if(a[i]<=c[1])
j = 1;
else if(a[i]>c[ans])
j = ++ans;
else
j = bin(ans,a[i]);
c[j] = a[i];
dp[i] = j;
}
return ans;
} int main()
{
int i;
while(~scanf("%d",&n))
{
for(i = 1; i<=n; i++)
scanf("%d",&a[i]);
printf("%d\n",LIS()); } return 0;
}
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