HDU 3530 RMQ+twopointer/单调队列

1. RMQ + 二分

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int n,m,k;
int a[100100];
int dp1[101000][20];
int dp2[101000][20];
void init()
{
for(int i=1; i<=n; i++) dp1[i][0] = dp2[i][0] = a[i];
for(int j=1; j<=20; j++)
for(int i=1; i+(1<<j)-1<=n; i++)
{
dp1[i][j] = max(dp1[i][j-1], dp1[i+(1<<(j-1))][j-1]);
dp2[i][j] = min(dp2[i][j-1], dp2[i+(1<<(j-1))][j-1]);
}
}
int get_max(int l, int r)
{
int k = log(r-l+1)/log(2);
return max(dp1[l][k], dp1[r-(1<<k)+1][k]);
}
int get_min(int l, int r)
{
int k = log(r-l+1)/log(2);
return min(dp2[l][k], dp2[r-(1<<k)+1][k]);
}
int ans = 0;
void solve(int i)
{
int l = i;
int r = n;
int res1;
while(l<=r)
{
int mid = (l+r)>>1;
int maxn = get_max(i, mid);
int minn = get_min(i, mid);
if(maxn - minn <=k)
{
res1 = mid;
l = mid+1;
}
else r = mid-1;
}
l = i;
r = n;
int res2;
while(l<=r)
{
int mid = (l+r)>>1;
int maxn = get_max(i, mid);
int minn = get_min(i, mid);
if(maxn - minn >= m)
{
res2 = mid;
r = mid-1;
}
else l = mid+1;
}
if(res1 >= res2)
ans = max(ans, res1 - res2 + 1);
}
int main()
{
while(~scanf("%d %d %d", &n, &m, &k))
{
memset(dp1, -1, sizeof(dp1));
memset(dp2, 0x3f3f3f3f, sizeof(dp2));
for(int i=1; i<=n; i++)
scanf("%d", &a[i]);
init();
ans = 0;
for(int i=1; i<=n; i++)
solve(i);
printf("%d\n", ans);
}
return 0;
}

2. RMQ + two pointer

#include <iostream>## 标题 ##
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int n,m,k;
int a[100100];
int dp1[101000][20];
int dp2[101000][20];
void init()
{
for(int i=1; i<=n; i++) dp1[i][0] = dp2[i][0] = a[i];
for(int j=1; j<=20; j++)
for(int i=1; i+(1<<j)-1<=n; i++)
{
dp1[i][j] = max(dp1[i][j-1], dp1[i+(1<<(j-1))][j-1]);
dp2[i][j] = min(dp2[i][j-1], dp2[i+(1<<(j-1))][j-1]);
}
}
int get_max(int l, int r)
{
int k = log2(r-l+1);
return max(dp1[l][k], dp1[r-(1<<k)+1][k]);
}
int get_min(int l, int r)
{
int k = log2(r-l+1);
return min(dp2[l][k], dp2[r-(1<<k)+1][k]);
}
int ans = 0;
void solve(int i)
{
int l = i;
int r = n;
int res1;
while(l<=r)
{
int mid = (l+r)>>1;
int maxn = get_max(i, mid);
int minn = get_min(i, mid);
if(maxn - minn <=k)
{
res1 = mid;
l = mid+1;
}
else r = mid-1;
}
l = i;
r = n;
int res2;
while(l<=r)
{
int mid = (l+r)>>1;
int maxn = get_max(i, mid);
int minn = get_min(i, mid);
if(maxn - minn >= m)
{
res2 = mid;
r = mid-1;
}
else l = mid+1;
}
if(res1 >= res2)
ans = max(ans, res1 - res2 + 1);
}
int main()
{
while(~scanf("%d %d %d", &n, &m, &k))
{
memset(dp1, -1, sizeof(dp1));
memset(dp2, 0x3f3f3f3f, sizeof(dp2));
for(int i=1; i<=n; i++)
scanf("%d", &a[i]);
init();
int l,r;
l=1;ans=0;           //l不用每次从1开始判，因为之前的l越小，r-l+1的值就越大。
for(int i=1;i<=n;i++)
{
r=i;
if(l>r)continue;
while(get_max(l,r)-get_min(l,r)>k)l++; //如果大于k,那么后面的r对于此时的l肯定不满足不大于k,所以必须l++;
if(get_max(l,r)-get_min(l,r)>=m && get_max(l,r)-get_min(l,r)<=k)ans=max(ans,r-l+1);
}
printf("%d\n",ans);
}
return 0;
}

3.单调队列

#include <bits/stdc++.h>
using namespace std;
int n,m,k;
int a[1000100];
deque<int> q1;
deque<int> q2;
int main()
{
while(cin>>n>>m>>k)
{
q1.clear(); q2.clear();
for(int i=1; i<=n; i++)
scanf("%d", &a[i]);
int now = 1;
int ans = 0;
for(int i=1; i<=n; i++)
{
while(!q1.empty()&&a[q1.back()]<a[i]) q1.pop_back();
while(!q2.empty()&&a[q2.back()]>a[i]) q2.pop_back();
q1.push_back(i); q2.push_back(i);
while(!q1.empty()&&!q2.empty()&&a[q1.front()] - a[q2.front()] > k)
{
if(q1.front() < q2.front())
{
now = q1.front()+1;
q1.pop_front();
}
else
{
now = q2.front()+1;
q2.pop_front();
}
}
if(!q1.empty()&&!q2.empty()&&a[q1.front()] - a[q2.front()] >= m)
ans = max(ans, i - now +1);
}
cout<<ans<<endl;
}
return 0;
}