hdu 5289 Assignment

传送门：

http://acm.hdu.edu.cn/showproblem.php?pid=5289

题意：找出一段区间里面最大值减去最小值小于k的区间个数

由于昨天做了那道线段树区间最值的题目，因此想起了这道多校第一场的第二题，用了双指针加线段树，秒A！

注意用线段树判断的时候别越界了！

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1e5+10;
int a[maxn];
int tree[3][4*maxn],n,k;

void build(int idx,int st,int en)
{
int l=idx<<1;
int r=l|1;
int mid=(st+en)>>1;

if(st==en)
{
tree[0][idx]=a[st];
tree[1][idx]=a[st];
return ;
}
build(l,st,mid);
build(r,mid+1,en);

tree[0][idx]=min(tree[0][l],tree[0][r]);
tree[1][idx]=max(tree[1][l],tree[1][r]);
}

int query(int idx,int st,int en,int i,int j,int ty)
{
int l=idx<<1;
int r=l|1;
int mid=(st+en)>>1;

if(st==i && en==j)
{
return tree[ty][idx];
}

if(j<=mid)return query(l,st,mid,i,j,ty);
else if(i>mid)return query(r,mid+1,en,i,j,ty);
else
{
if(ty==0)return min(query(l,st,mid,i,mid,ty),query(r,mid+1,en,mid+1,j,ty));
else return max(query(l,st,mid,i,mid,ty),query(r,mid+1,en,mid+1,j,ty));
}
}

int main(){
int t;cin>>t;
while(t--){
ll sum=0;
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
sum=n;
build(1,1,n);int r=2;
for(int l=1;l<n;l++){
while(r<=n&&query(1,1,n,l,r,1)-query(1,1,n,l,r,0)<k){
sum+=r-l;r++;
}

}
printf("%lld\n",sum);
}
}

又学习了单调队列的做法，着实是比较简便，可以O（1）的查询给定区间的最大最小值，再利用双指针滑动就ok了！

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std ;
#define LL long long
deque <LL> Max , Min ;
//单调队列，Max最大值，Min最小值
LL a[100010] ;
int main()
{
int T , n , i , j ;
LL k , ans ;
scanf("%d", &T) ;
while( T-- )
{
scanf("%d %I64d", &n, &k) ;
for(i = 0 ; i < n ; i++)
scanf("%I64d", &a[i]) ;
while( !Max.empty() ) Max.pop_back() ;
while( !Min.empty() ) Min.pop_back() ;
for(i = 0 , j = 0 , ans = 0; i < n ; i++)  //i在前，j在后
{
while( !Max.empty() && Max.back() < a[i] ) Max.pop_back() ;
Max.push_back(a[i]) ;
while( !Min.empty() && Min.back() > a[i] ) Min.pop_back() ;
Min.push_back(a[i]) ;
while( !Max.empty() && !Min.empty() && Max.front() - Min.front() >= k )
{
ans += (i-j) ;
if( Max.front() == a[j] ) Max.pop_front() ;
if( Min.front() == a[j] ) Min.pop_front() ;
j++ ;
}
}
while( j < n )
{
ans += (i-j) ;
j++ ;
}
printf("%lld\n", ans) ;
}
return 0 ;
}

更有rmq加二分的写法，这个是新的二分写法

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
using namespace std;

int  maxsum[100000][30];
int minsum[100000][30];

int a[100000];
int n,k;
void rmq_init()
{
for(int j = 1; (1<<j) <= n; ++j)
for(int i = 1; i + (1<<j) - 1 <= n; ++i)
{
maxsum[i][j] = max(maxsum[i][j-1],maxsum[i+(1<<(j-1))][j-1]);
minsum[i][j] = min(minsum[i][j-1],minsum[i+(1<<(j-1))][j-1]);
}

}

int query(int l, int r)
{
int k = log2(r-l+1);
int Max = max(maxsum[l][k], maxsum[r-(1<<k)+1][k]);
int Min = min(minsum[l][k], minsum[r-(1<<k)+1][k]);
return Max - Min;
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
for(int i = 1; i <= n;++i)
{
scanf("%d",a+i);
maxsum[i][0] = minsum[i][0] = a[i];
}
rmq_init();

long long  ans = 0;
int l , r;
for(int i = 1; i <= n; ++i)
{
l = i , r = n;

while(l <= r)
{
int mid = (l+r)/2;
int cha = query(i,mid);
if(cha < k) l = mid+1;
else r = mid - 1;
}

ans += l - i;
}

printf("%lld\n",ans);

}
return 0;
}