poj 2533 & poj 1631 Longest Ordered Subsequence( LIS果题 )
2014-09-02 08:53
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题目链接:
POJ 2533:http://poj.org/problem?id=2533
POJ 1631:http://poj.org/problem?id=1631
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
Sample Output
Source
Northeastern Europe 2002, Far-Eastern Subregion
LIS果题。
代码如下:
POJ 1631 和这题基本上一样,只需改一下数组大小,叫一个多测试案例输入的循环即可!
这里也贴一下代码:
POJ 2533:http://poj.org/problem?id=2533
POJ 1631:http://poj.org/problem?id=1631
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
Source
Northeastern Europe 2002, Far-Eastern Subregion
LIS果题。
代码如下:
#include <cstdio> #include <iostream> #include <algorithm> int N; int ans; int a[1017], dp[1017]; int bin(int len, int tem) { int l = 1, r = len; while(l <= r) { int mid = (l+r)/2; if(tem > dp[mid]) l = mid+1; else r = mid-1; } return l; } int LIS(int *b) { dp[1] = a[1]; ans = 1; int k; for(int i = 2; i <= N; i++) { if(a[i] < dp[1]) k = 1; else if(a[i] > dp[ans]) k = ++ans; else k = bin(ans,a[i]); dp[k] = a[i]; } return ans; } int main() { while(~scanf("%d",&N)) { for(int i = 1; i <= N; i++) { scanf("%d",&a[i]); } LIS(a); printf("%d\n",ans); } return 0; }
POJ 1631 和这题基本上一样,只需改一下数组大小,叫一个多测试案例输入的循环即可!
这里也贴一下代码:
#include <cstdio> #include <iostream> #include <algorithm> int N; int ans; int a[40017], dp[40017]; int bin(int len, int tem) { int l = 1, r = len; while(l <= r) { int mid = (l+r)/2; if(tem > dp[mid]) l = mid+1; else r = mid-1; } return l; } int LIS(int *b) { dp[1] = a[1]; ans = 1; int k; for(int i = 2; i <= N; i++) { if(a[i] < dp[1]) k = 1; else if(a[i] > dp[ans]) k = ++ans; else k = bin(ans,a[i]); dp[k] = a[i]; } return ans; } int main() { int t; scanf("%d",&t); while(t--) { scanf("%d",&N); for(int i = 1; i <= N; i++) { scanf("%d",&a[i]); } LIS(a); printf("%d\n",ans); } return 0; }
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