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POJ 2533 Longest Ordered Subsequence(LIS模版题)
Longest Ordered Subsequence
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 47465 | Accepted: 21120 |
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
71 7 3 5 9 4 8
Sample Output
4
题目链接:POJ 2533
LIS模版题,N2和N*logN两种写法
N2代码:
#include<stdio.h>#include<iostream>#include<algorithm>#include<cstdlib>#include<sstream>#include<cstring>#include<bitset>#include<string>#include<deque>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>using namespace std;#define INF 0x3f3f3f3f#define CLR(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=1e3+10;int arr[N],mx[N];void init(){ CLR(arr,0); CLR(mx,0);}int main(void){ int n,i,j,pre_len,next_len; while (~scanf("%d",&n)) { init(); for (i=1; i<=n; ++i) scanf("%d",&arr[i]); mx[1]=1; for (i=2; i<=n; ++i) { pre_len=0; for (j=1; j<i; ++j) { if(arr[j]<arr[i])//arr[i]可以接到arr[j]后面 if(mx[j]>pre_len)//接到一个具有最长LIS的后面。 pre_len=mx[j]; } mx[i]=pre_len+1; } printf("%d\n",*max_element(mx+1,mx+1+n)); } return 0;}
NlogN代码:
#include<stdio.h>#include<iostream>#include<algorithm>#include<cstdlib>#include<sstream>#include<cstring>#include<bitset>#include<string>#include<deque>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>using namespace std;#define INF 0x3f3f3f3f#define CLR(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=1e3+10;int arr[N],d[N];void init(){ CLR(arr,0); CLR(d,0);}int main(void){ int n,i,j,mxlen; while (~scanf("%d",&n)) { init(); for (i=1; i<=n; ++i) scanf("%d",&arr[i]); mxlen=1; d[mxlen]=arr[mxlen]; for (i=2; i<=n; ++i) { if(d[mxlen]<arr[i]) d[++mxlen]=arr[i];//最好情况一直往后增长 else { int pos=lower_bound(d,d+mxlen,arr[i])-d;//用二分找到一个下界可放置位置 d[pos]=arr[i]; } } printf("%d\n",mxlen); } return 0;}
POJ 2533 Longest Ordered Subsequence(LIS模版题)
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