POJ2533:Longest Ordered Subsequence(LIS) 动态规划
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题意:输出最长递增子序列长度
思路一:刷表法
#include<cstdio>#include<cstring>#include <iostream>#define maxn 1005using namespace std;int a[maxn];int dp[maxn];int main(){ int n,ans=0; scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%d",&a[i]); dp[i]=1; } //DP,从第一个元素开始,如果这个元素后面有a[j]>a[i],且dp[j]<dp[i]+1,那么更新dp[j] for(int i=0;i<n-1;i++) for(int j=i+1;j<n;j++) if(a[j]>a[i]&&dp[j]<=dp[i]) dp[j]=dp[i]+1; for(int i=0;i<n;i++){//找出最大长度 if(dp[i]>ans) ans=dp[i]; } printf("%d\n",ans); return 0;}
思路二:填表法
#include<cstdio>#include<cstring>#include <iostream>#define maxn 1005using namespace std;int a[maxn];int dp[maxn];int main(){ int n,ans=0; scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%d",&a[i]); dp[i]=1; }//DP从第一个元素开始,找到该元素能接上的最大长度并记录 for(int i=0;i<n;i++){ int flag=0; for(int j=0;j<i;j++){ if(a[j]<a[i]&&dp[j]>flag) flag=dp[j];//a[i]能够接上的最长序列 } dp[i]=flag+1; } for(int i=0;i<n;i++){ if(dp[i]>ans) ans=dp[i]; } printf("%d\n",ans); return 0;}
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