Almost Sorted Array HDU

Almost Sorted Array HDU - 5532

题意：

如果一个数组的元素“按非递减或非递增顺序排列”，那么称这个数组有序。现在，LYD给了你一个数组a，他让你从中选择并移除仅仅1个元素，请问剩下的元素构成的数组有可能是有序的吗？

题解：

其实就是让你求一个数组的最长上升或者下降下序列，这个序列是否大于等于n-1，这样多余的数的数量才会小于等于1，直接用nlogn的方法求LIS
或者直接模拟写法，我们同时判断一个数列的上升和下降情况，上升时，下降的情况不能出现多于一次，下降情况时，上升的情况不能多于一次。
两个方法详细看代码

代码：

nlogn求LIS

#include<bits/stdc++.h>
#define debug(a,b) printf("%s = %d\n",a,b);
typedef long long ll;
using namespace std;inline int read(){int s=0,w=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();//s=(s<<3)+(s<<1)+(ch^48);return s*w;
}
const int maxn=1e5+9;
int dp[maxn];
int val[maxn];
int main()
{int t;cin>>t;while(t--){memset(dp,0,sizeof(dp));int n;int cnt1=0,cnt2=0;n=read();int temp;for(int i=1;i<=n;i++)val[i]=read();for(int i=1;i<=n;i++){//求单调上升序列 if(val[i]>=dp[cnt1])dp[++cnt1]=val[i];else {temp=upper_bound(dp+1,dp+1+cnt1,val[i])-dp;dp[temp]=val[i];}}memset(dp,0,sizeof(dp));for(int i=n;i>=1;i--){//求单调递减序列 if(val[i]>=dp[cnt2])dp[++cnt2]=val[i];else {temp=upper_bound(dp+1,dp+1+cnt2,val[i])-dp;dp[temp]=val[i];}}if(cnt1>=n-1||cnt2>=n-1)printf("YES\n");else printf("NO\n");}return 0;
}

模拟做法

#include<stdio.h>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn=100005;
int n,sor[maxn];
bool judge1()//第一次非递减
{int sum=0;sor[0]=-inf;sor[n+1]=inf;for(int i=2;i<=n;i++){if(sor[i]<sor[i-1]){if(sum==1)return 0;sum++;if(sor[i+1]>=sor[i-1]||sor[i]>=sor[i-2])continue;elsereturn 0;}}return 1;
}
bool judge2()//第二次非递增
{int sum=0;sor[0]=inf;sor[n+1]=-inf;for(int i=2;i<=n;i++){if(sor[i]>sor[i-1]){if(sum==1)return 0;sum++;if(sor[i+1]<=sor[i-1]||sor[i]<=sor[i-2])continue;elsereturn 0;}}return 1;
}
int main()
{int t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&sor[i]);if(judge1()||judge2())printf("YES\n");elseprintf("NO\n");}return 0;
}

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Almost Sorted Array HDU - 5532

题意：

题解：

代码：

nlogn求LIS

模拟做法

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