HDU1087 Super Jumping! Jumping! Jumping! 动态规划
2014-10-31 12:24
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动态规划
本题其实就是求一个序列的最长上升子序列,状态转移方程为:dp【i】=max(dp【i】,dp【j】+ans【i】)
本题其实就是求一个序列的最长上升子序列,状态转移方程为:dp【i】=max(dp【i】,dp【j】+ans【i】)
#include <cstdio> #include <iostream> #define MAX 1100 using namespace std; int main() { int ans[MAX],dp[MAX]; int n,i; while(scanf("%d",&n)&&n) { for(i=0;i<n;i++) scanf("%d",&ans[i]); for(i=0;i<n;i++) dp[i]=ans[i]; int temp=ans[0]; for(i=0;i<n;i++) { for(int j=0;j<=i;j++) if(ans[i]>ans[j]) dp[i]=max(dp[i],dp[j]+ans[i]); temp=max(temp,dp[i]); } printf("%d\n",temp); } return 0; }
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