POJ 2533 Longest Ordered Subsequence

Description

A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 >= i1 < i2 < ... < iK >= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
Input

The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 >= N >= 1000
Output

Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
Sample Input

7
1 7 3 5 9 4 8
Sample Output

4

dd[1] = 1;
dd[k] = Max (dd[i]:1 >= i < k 且 df[i ]< df[k] 且 k != 1) + 1.

n：7
i ：0 1 2 3 4 5 6
df ：1 7 3 5 9 4 8
dd[0]:1;
dd[1]:dd[0]+1=2;
dd[2]:dd[0]+1=2;
dd[3]:dd[2]+1=3;
dd[4]:因为df[0],df[1],df[2],df[3]都小于df[4]，但是dd[3]最大，

dd[5]:dd[2]+1=3;
dd[6]:dd[5]+1=4;
..................................................................

```#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define Maxn 1020
int df[Maxn],dd[Maxn];

int cmp(const void *x,const void *y){
return (*(int *)y-*(int *)x);
/*快速排序，从大到小排序*/
}
int main(){
int n;
while(scanf("%d",&n)==1){
for(int i=0;i<n;i++){
scanf("%d",&df[i]);
}
dd[0]=1;
for(int i=1;i<n;i++){
int t=0;
for(int j=0;j<i;j++){
if(df[i]>df[j]){
if(t<dd[j]){
t=dd[j];
}
}
}
dd[i]=t+1;
//此时的t是dd[i]之前的最大增子序列的个数
}
qsort(dd,n,sizeof(int),cmp);
printf("%d\n",dd[0]);
}
return 0;
}

```
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