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Wavio is a sequence of integers. It has some interestingproperties.
· Wavio is of odd length i.e. L = 2*n+ 1.
· The first (n+1) integers ofWavio sequence makes a strictly increasing sequence.
· The last (n+1) integers of Waviosequence makes a strictly decreasing sequence.
· No two adjacent integers are same in aWavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Waviosequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not avalid wavio sequence. In this problem, you will be given a sequence ofintegers. You have to find out the length of the longest Wavio sequence whichis a subsequence of the given sequence. Consider, the given sequence as :
1 2 3 2 1 2 3 4 32 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 outputwill be 9.
Input
The input filecontains less than 75 test cases. The description of each test case isgiven below: Input is terminated by end of file.
Each set starts with a postiveinteger, N(1<=N<=10000). In next few lines there will be Nintegers.
Output
For each set of input print the length of longest waviosequence 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
|
这道题主要是LIS的应用,把串正反走一遍就行,代码中LIS()函数是精华。
代码实现:
#include#include using namespace std;#define N 10001int dp1[10001],dp2[10001],a[10001],b[10001];int t;void LIS(int dp[],int a[]){ int stack[N]; int top=0; stack[top]=-99999999; for(int i=1; i<=t; i++) { //如果a[i]>栈顶部元素,则压栈 if(a[i]>stack[top]) { stack[++top]=a[i]; dp[i]=top; } //如果a[i]不大于栈顶部元素,则二分查找第一个比a[i]大的元素 else { int l=1,r=top; while(l<=r) { int mid=(l+r)>>1; if(a[i]>stack[mid]) { l=mid+1; } else r=mid-1; } //替换a[i] stack[l]=a[i]; dp[i]=l; } }}int main(){ while(cin>>t) { for(int i=1;i<=t;i++) { cin>>a[i]; b[t-i+1]=a[i]; } memset(dp1,0,sizeof(dp1)); memset(dp2,0,sizeof(dp2)); LIS(dp1,a); LIS(dp2,b); int maxx=-1,ans=0; for(int i=1;i<=t;i++) { ans=min(dp1[i],dp2[t-i+1])*2-1; if(ans>maxx) { maxx=ans; } } cout< <
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