POJ-2533最長上升子序列(DP+二分）（最佳化版）

標籤：

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 41944		Accepted: 18453

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

71 7 3 5 9 4 8

Sample Output

4

Source

Northeastern Europe 2002, Far-Eastern Subregion 方法一：記憶化搜尋缺點：時間複雜度O(n^2)

#include <cstdio>#include <iostream>#include <cstdlib>#include <algorithm>#include <ctime>#include <cmath>#include <string>#include <cstring>#include <stack>#include <queue>#include <list>#include <vector>#include <map>#include <set>using namespace std;const int INF=0x3f3f3f3f;const double eps=1e-10;const double PI=acos(-1.0);#define maxn 1100int a[maxn];int dp[maxn];int dfs(int p){    if(dp[p] != -1) return dp[p];    int res = 0;    for(int i = 0; i < p; i++)        if(a[p] > a[i])            res = max(res, dfs(i)+1);    dp[p] = res;    return res;}int main(){    int n;    while(~scanf("%d", &n))    {        memset(dp, -1, sizeof dp);        for(int i = 0; i < n; i++)            scanf("%d", &a[i]);        int pp = -1;        for(int j = 0;  j < n; j++)        {            pp = max(pp, dfs(j)+1);        }            //printf("%d\n", dfs(n-1)+ 1);            printf("%d\n", pp);    }    return 0;}

方法二：dp+二分

其中low_bound 返回第一個大於它的數的下標。

缺點：無法儲存每個以 a[i]結尾的最長上升子序列。

#include <cstdio>#include <iostream>#include <cstdlib>#include <algorithm>#include <ctime>#include <cmath>#include <string>#include <cstring>#include <stack>#include <queue>#include <list>#include <vector>#include <map>#include <set>using namespace std;const int INF=0x3f3f3f3f;const double eps=1e-10;const double PI=acos(-1.0);#define maxn 11000int a[maxn];int dp[maxn];int main(){    int n;    while(~scanf("%d", &n))    {        for(int i = 0; i < n; i++)            scanf("%d", &a[i]);        int cnt = 0;        //memset(dp, INF, sizeof dp);        dp[cnt] = a[0];        for(int i = 1; i < n; i++)        {            if(a[i] > dp[cnt])            {                dp[++cnt] = a[i];            }            else            {                int pos = lower_bound(dp,dp+cnt+1,a[i]) - dp;                dp[pos] = a[i];            }        }        printf("%d\n", cnt+1);    }    return 0;}

方法三：dp+二分（最佳化版）

彌補了上面兩種方法不足。時間複雜度為O（nlogn) 又能儲存每個以a[i]結尾的最長上升子序列。

#include <cstdio>#include <iostream>#include <cstdlib>#include <algorithm>#include <ctime>#include <cmath>#include <string>#include <cstring>#include <stack>#include <queue>#include <list>#include <vector>#include <map>#include <set>using namespace std;const int INF=0x3f3f3f3f;const double eps=1e-10;const double PI=acos(-1.0);#define maxn 11000int a[maxn];int b[maxn];int dp[maxn];int main(){    int n;    while(~scanf("%d", &n))    {        for(int i = 0; i < n; i++)            scanf("%d", &a[i]);        memset(dp, 0, sizeof dp);        memset(b, INF, sizeof b);        for(int i = 0; i < n; i++)        {            int pos = lower_bound(b,b+n,a[i]) - b;            dp[i] = pos+1;            b[pos] = a[i];        }        int ans = -1;        for(int  i = 0; i < n; i++)            ans = max(ans, dp[i]);        printf("%d\n", ans);    }    return 0;}

 

POJ-2533最長上升子序列(DP+二分）（最佳化版）