UVA 10534 Wavio Sequence (波浪DP,4级)
Description
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Problem D
Wavio Sequence
Input: Standard Input
Output: Standard Output
Time Limit: 2 Seconds
Wavio is a sequence of integers. It has some interesting properties.
· Wavio is of odd length i.e. L = 2*n + 1.
· The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.
· The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.
· No two adjacent integers are same in a Wavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider, the given sequence as :
1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.
Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be9.
Input
The input file contains less than 75 test cases. The description of each test case is given below: Input is terminated by end of file.
Each set starts with a postive integer, N(1<=N<=10000). In next few lines there will beN integers.
Output
For each set of input print the length of longest wavio sequence in a line.
Sample Input Output for Sample Input
10 1 2 3 4 5 4 3 2 1 10 19 1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1 5 1 2 3 4 5 |
9 9 1
|
Problemsetter: Md. Kamruzzaman, Member of Elite Problemsetters' Panel
求最长的子序列从左到中递增,中到有递减,
思路:字符串取反,求个递增序定中间点。求最优
#include<cstring> #include<cstdio> #include<iostream> #define FOR(i,a,b) for(int i=a;i<=b;++i) #define clr(f,z) memset(f,z,sizeof(f)) using namespace std; const int mm=10009; int f[mm],t[mm]; int idp[mm],ddp[mm]; int main() { int n; while(~scanf("%d",&n)) { FOR(i,1,n)scanf("%d",&f[i]); clr(t,0x3f);clr(idp,0); FOR(i,1,n) { int x=lower_bound(t+1,t+n+1,f[i])-t; idp[i]=max(idp[i],x); t[x]=f[i]; //cout<<i<<" "<<idp[i]<<endl; } clr(t,0x3f);clr(ddp,0); int ans=0; for(int i=n;i>=1;--i) { int x=lower_bound(t+1,t+n+1,f[i])-t; ddp[i]=max(ddp[i],x); t[x]=f[i]; //cout<<i<<" "<<ddp[i]<<endl; ans=max(ans,min(idp[i],ddp[i])); } ans=ans+ans-1; printf("%d\n",ans); } return 0; }