UVa 10534 Wavio Sequence (最长递增子序列 DP 二分)
Wavio Sequence
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 length 9. 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 be 9.
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 be N 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
题意 求一个序列a某一位的最长递增序列(lis)和最长递减序列(lds)中最小值的最大值
開始直接用DP写了 然后就超时了 后来看到别人说要用二分把时间复杂度优化到O(n*logn) 果然如此
用一个栈s保存长度为i的LIS的最小尾部s[i] top为栈顶即当前LIS的长度 初始s[1]=a[1] top=1 遍历整个序列 当a[i]>s[top]时 a[i]入栈 in[i]=top
否则 在栈中查找（二分）第一个大于等于a[i]的下标pos 并替换 这样就添加了LIS增长的潜力 in[i]=pos;
in[i]表示以a[i]为尾部的LIS的长度；
求lds的过程也是类似
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 10005;
int a[N], in[N], de[N], s[N], m, n, top, pos;
int BinSearch (int k, int le, int ri)
{
while (le <= ri)
{
m = (le + ri) >> 1;
if (s[m] >= k) ri = m - 1;
else le = m + 1;
}
return ri + 1;
}
void lis()
{
memset (s, 0, sizeof (s));
memset (in, 0, sizeof (in));
s[1] = a[1];
in[1] = top = 1;
for (int i = 2; i <= n; ++i)
{
if (s[top] < a[i])
{
s[++top] = a[i];
in[i] = top;
}
else
{
pos = BinSearch (a[i], 1, top);
s[pos] = a[i];
in[i] = pos;
}
}
}
void lds()
{
memset (s, 0, sizeof (s));
memset (de, 0, sizeof (de));
s[1] = a[n];
de[n] = top = 1;
for (int i = n - 1; i >= 1; --i)
{
if (s[top] < a[i])
{
s[++top] = a[i];
de[i] = top;
}
else
{
pos = BinSearch (a[i], 1, top);
s[pos] = a[i];
de[i] = pos;
}
}
}
int main()
{
while (scanf ("%d", &n) != EOF)
{
for (int i = 1; i <= n; ++i)
scanf ("%d", &a[i]);
int ans = 1;
lis();
lds();
for (int i = 1; i <= n; ++i)
{
if (min (de[i], in[i]) > ans)
ans = min (de[i], in[i]);
}
printf ("%d\n", ans * 2 - 1);
}
return 0;
}
还有TLE的DP版
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=10005;
int a[N],c[N],d[N],n;
int dpde(int i)
{
if(d[i]) return d[i];
d[i]=1;
for(int j=i;j<=n;++j)
{
if(a[i]>a[j])
d[i]=max(d[i],dpde(j)+1);
}
return d[i];
}
int dpin(int i)
{
if(c[i]) return c[i];
c[i]=1;
for(int j=i;j>=1;--j)
{
if(a[i]>a[j])
c[i]=max(c[i],dpin(j)+1);
}
return c[i];
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
int ans=1;
memset(d,0,sizeof(d));
memset(c,0,sizeof(c));
for(int i=1;i<=n;++i)
{
if(min(dpde(i),dpin(i))>ans)
ans=min(d[i],c[i]);
}
printf("%d\n",ans*2-1);
}
return 0;
}
