poj 2533 Longest Ordered Subsequence
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题目来源:POJ 2533
思路:
裸的最长严格上升子序列。
动态转移方程:dp[j]=max(dp[i]+1,dp[j]),其中i<j,temp[i]<temp[j]
其中dp[j]是指以下标j元素为结尾的LIS的长度,temp数组存储的是整个序列。
#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string.h>using namespace std;const int MAXN=1000+5;int temp[MAXN];int dp[MAXN];int main(){ int N; scanf("%d",&N); int i,j; for(i=1;i<=N;++i) scanf("%d",&temp[i]); for(i=1;i<=N;++i) dp[i]=1; for(i=2;i<=N;++i) for(j=1;j<i;++j) { if(temp[i]>temp[j]) dp[i]=max(dp[j]+1,dp[i]); } sort(dp+1,dp+1+N); printf("%d\n",dp[N]); return 0;}
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