Codeforces Round #333 (Div. 2)B. Approximating a Constant Range(ST+二分||set暴力)

题目链接

题意：给你n个数字，然后问你，最长区间是多长（区间满足最大减去最小<=1）

解法：二分每个点往右边可以扩展多少距离，区间最大最小用ST算法

#define CF

#ifndef CF
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#else
#include<bits/stdc++.h>
#endif // CF
using namespace std;
#define LL long long
#define pb push_back
#define X first
#define Y second
#define cl(a,b) memset(a,b,sizeof(a))
typedef pair<long long ,long long > P;
const int maxn=100005;
const int inf=1<<29;
const LL mod=1e9+7;
int ma[maxn][30],mi[maxn][30];

void init_rmq(int n){//初始化ST
for(int j=1;j<20;j++){
for(int i=1;i+(1<<j)-1<=n;i++){
ma[i][j]=max(ma[i][j-1],ma[i+(1<<(j-1))][j-1]);
mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
}
}
}
int query(int l,int r){//查询最大值，和最小值
if(l>r)swap(l,r);
int k=(int)(log(r-l+1)/log(2.0));
int a=max(ma[l][k],ma[r-(1<<k)+1][k]);
int b=min(mi[l][k],mi[r-(1<<k)+1][k]);
return a-b;
}
int bs(int l,int r){///二分每个点往右能跑多远
int m;int ll=l,rr=r;
while(l+1<r){
m=(l+r)/2;
if(query(ll,m)<=1)l=m;
else r=m;
}
int ans=l-ll+1;
if(query(ll,rr)<=1){
ans=max(ans,rr-ll+1);
}
return ans;
}
int main(){
int n;scanf("%d",&n);
for(int i=1;i<=n;i++){
int x;scanf("%d",&x);
mi[i][0]=ma[i][0]=x;
}
init_rmq(n);
int ans=0;
for(int i=1;i<=n;i++){
ans=max(ans,bs(i,n));
}
printf("%d\n",ans);
return 0;
}

第二种解法：直接使用multiset进行暴力，思路是以i点结尾的话能够跑多远

#define CF

#ifndef CF
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#else
#include<bits/stdc++.h>
#endif // CF
using namespace std;
#define LL long long
#define pb push_back
#define X first
#define Y second
#define cl(a,b) memset(a,b,sizeof(a))
typedef pair<long long ,long long > P;
const int maxn=100005;
const int inf=1<<29;
const LL mod=1e9+7;
multiset<int>s;
int a[maxn];
int main(){
int n;scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
int pos=0,ans=0;
for(int i=0;i<n;i++){
s.insert(a[i]);
while(*s.rbegin()-*s.begin()>1){
s.erase(s.find(a[pos++]));
}
ans=max(ans,i-pos+1);
}
printf("%d\n",ans);
return 0;
}