UVA-1608 Non-boring sequences
2017-10-14 13:39
375 查看
题目链接:https://vjudge.net/problem/UVA-1608
先进行一次预处理,保存每个元素左边和右边最近元素的位置,可以在O(1)时间内判断出某个元素是否唯一
再用分治法检查每个子序列是否non-boring,时间复杂度O(nlogn)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<string>
using namespace std;
typedef long long ll;
const int N=2*1e5+10;
int l
,r
;
int a
,n;
bool find(int d,int u)
{
if(d>=u) return true;
int d0=d,u0=u;
while(d0<=u0)
{
if(l[d0]<d&&r[d0]>u)
return find(d,d0-1)&&find(d0+1,u);
if(l[u0]<d&&r[u0]>u)
return find(d,u0-1)&&find(u0+1,u);
d0++;
u0--;
}
return false;
}
int main()
{
//freopen("/home/zlwang/Desktop/t2.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
memset(l,-1,sizeof(l));
for(int i=0;i<n;i++)
r[i]=n;
map<int,int> m;
for(int i=0;i<n;i++)
{
if(!m.count(a[i])) m[a[i]]=i;
else
{
l[i]=m[a[i]];
r[m[a[i]]]=i;
m[a[i]]=i;
}
}
if(find(0,n-1)) printf("non-boring\n");
else printf("boring\n");
}
return 0;
}
先进行一次预处理,保存每个元素左边和右边最近元素的位置,可以在O(1)时间内判断出某个元素是否唯一
再用分治法检查每个子序列是否non-boring,时间复杂度O(nlogn)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<string>
using namespace std;
typedef long long ll;
const int N=2*1e5+10;
int l
,r
;
int a
,n;
bool find(int d,int u)
{
if(d>=u) return true;
int d0=d,u0=u;
while(d0<=u0)
{
if(l[d0]<d&&r[d0]>u)
return find(d,d0-1)&&find(d0+1,u);
if(l[u0]<d&&r[u0]>u)
return find(d,u0-1)&&find(u0+1,u);
d0++;
u0--;
}
return false;
}
int main()
{
//freopen("/home/zlwang/Desktop/t2.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
memset(l,-1,sizeof(l));
for(int i=0;i<n;i++)
r[i]=n;
map<int,int> m;
for(int i=0;i<n;i++)
{
if(!m.count(a[i])) m[a[i]]=i;
else
{
l[i]=m[a[i]];
r[m[a[i]]]=i;
m[a[i]]=i;
}
}
if(find(0,n-1)) printf("non-boring\n");
else printf("boring\n");
}
return 0;
}
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