codeforces#333 div2 B. Approximating a Constant Range
http://codeforces.com/contest/602/problem/B
这道题的两种做法,
滑窗+线段树查询区间最值:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) using namespace std; typedef long long ll; const int maxn=1000100; const int INF=(1<<29); struct SegTree { struct Node { int l,r; int Min,Max; }; Node T[maxn<<2]; void push_up(int rt) { T[rt].Max=max(T[rt<<1].Max,T[rt<<1|1].Max); T[rt].Min=min(T[rt<<1].Min,T[rt<<1|1].Min); } void build(int *a,int l,int r,int rt) { T[rt].l=l;T[rt].r=r; if(l==r){ T[rt].Max=T[rt].Min=a[l]; return; } int m=(l+r)>>1; build(a,l,m,rt<<1); build(a,m+1,r,rt<<1|1); push_up(rt); } int queryMax(int L,int R,int rt) { if(L<=T[rt].l&&T[rt].r<=R) return T[rt].Max; int m=(T[rt].l+T[rt].r)>>1; int res=-INF; if(L<=m) res=max(res,queryMax(L,R,rt<<1)); if(R>m) res=max(res,queryMax(L,R,rt<<1|1)); return res; } int queryMin(int L,int R,int rt) { if(L<=T[rt].l&&T[rt].r<=R) return T[rt].Min; int m=(T[rt].l+T[rt].r)>>1; int res=INF; if(L<=m) res=min(res,queryMin(L,R,rt<<1)); if(R>m) res=min(res,queryMin(L,R,rt<<1|1)); return res; } };SegTree sgt; int a[maxn],n; int main() { //freopen("in.txt","r",stdin); while(cin>>n){ REP(i,1,n) scanf("%d",&a[i]); sgt.build(a,1,n,1); int pre=1; int ans=1; REP(i,1,n){ while(sgt.queryMax(pre,i,1)-sgt.queryMin(pre,i,1)>1) pre++; ans=max(ans,i-pre+1); } cout<<ans<<endl; } return 0; }
滑窗+单调队列维护滑窗最值:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> #define REP(i,a,b) for(int i=a;i<=b;i++) #define MS0(a) memset(a,0,sizeof(a)) using namespace std; typedef long long ll; const int maxn=1000100; const int INF=1<<29; deque<int> q1,q2; int n,a[maxn]; int main() { freopen("in.txt","r",stdin); while(cin>>n){ REP(i,1,n) scanf("%d",&a[i]); int pre=1,ans=1; while(!q1.empty()) q1.pop_back(); while(!q2.empty()) q2.pop_back(); REP(i,1,n){ while(!q1.empty()&&q1.back()<a[i]) q1.pop_back(); q1.push_back(a[i]); while(!q2.empty()&&q2.back()>a[i]) q2.pop_back(); q2.push_back(a[i]); while(q1.front()-q2.front()>1){ if(a[pre]==q1.front()) q1.pop_front(); if(a[pre]==q2.front()) q2.pop_front(); pre++; } ans=max(ans,i-pre+1); } cout<<ans<<endl; } return 0; }
第一道单调队列= =
没有AC不了的题,只有不努力的ACMER!