POJ 2533--Longest Ordered Subsequence【最长递增子序列 + 二分优化】
Longest Ordered Subsequence
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 36560 | Accepted: 16087 |
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.
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
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[1100];
int dp[1100];
int main (){
int n,ans;
while(cin>>n){
int i, j;
for(i = 1;i <= n; ++i)
cin>>a[i];
ans = 1;
for(i=1;i<=n;++i)
dp[i]=1;
for(i = 2; i <= n; ++i)
for(j = i-1; j >= 1; --j){
if(a[i] > a[j] && dp[i] <= dp[j]){
dp[i] = dp[j] + 1;
if(dp[i] > ans) ans = dp[i];
}
}
cout<<ans<<endl;
}
return 0;
}
2015 年 4 月25更新
二分搜索，时间更快，还能记录最优解。
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MAX = 10005;
//二分法搜索digit，若s中存在digit，返回其下标
//若不存在，返回str中比digit小的最大那个数的（下标 + 1）
int serch(int s[], int digit,int length){
int l = 0, r = length, mid;
while(r != l){
mid = (l + r) / 2;
if(digit == s[mid])
return mid;
else if(digit < s[mid])
r = mid;
else
l = mid + 1;
}
return l;
}
int main (){
int n;
int i,j;
while(~scanf("%d", &n)){
int a[MAX];
int s[MAX];
for(i = 1; i <= n; ++i)
scanf("%d", &a[i]);
s[0] = -inf; //下界无穷小
int len = 1; //s的长度
for(i = 1; i <= n; ++i){
s[len] = inf;//上界无穷大，指针len总总指向s最后一个元素的后一位
j = serch(s, a[i], len);
//printf("----%d\n",j);
if(j == len) //a[i]大于s最大（最后）的元素
len++;
s[j] = a[i];
}
printf("%d\n", len - 1);//len 要减去1
//for(i = 1; i < len; ++i)
// printf("%d ",s[i]);
}
return 0;
}