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Assignment

2
4 2
3 1 2 4
10 5
0 3 4 5 2 1 6 7 8 9

5
28
Hint
First Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3] 
 

rmq+二分

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;
const int mod=1e9+7;

using namespace std;

#define N 100007

int a[N],n;
int dpmin[N][25],dpmax[N][25];
int k;

inline bool judge(int le,int ri)
{
    int kk=log2((ri-le+1)*1.0);
    int mi=min(dpmin[le][kk],dpmin[ri-(1<<kk)+1][kk]);
    int ma=max(dpmax[le][kk],dpmax[ri-(1<<kk)+1][kk]);
    return ma-mi<k;
}

int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
            dpmin[i][0]=dpmax[i][0]=a[i];

        for(j=1;(1<<j)<=n;j++)
            for(i=1;i+(1<<j)-1<=n;i++)
            {
                int p=1<<(j-1);
                dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+p][j-1]);
                dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+p][j-1]);
            }

         __int64 ans=0;
         int le,ri,p;
         for(i=1;i<=n;i++)
         {
             le=i;
             ri=n;
             while(le<=ri)
             {
                 int mid=(le+ri)>>1;
                 if(judge(i,mid))
                 {
                     p=mid;
                     le=mid+1;
                 }
                 else
                    ri=mid-1;
             }
             ans+=p-i+1;
         }
         printf("%I64d\n",ans);
    }
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8

using namespace std;
const int mod=1e9+7;

#define INF 0x3f3f3f3f

const int N=100005;

int mique[N],maque[N],mihead,mahead,mitail,matail;
int n,a[N],k;

int pre,now;
__int64 ans;

inline void miinque(int i)
{
    while(mihead<mitail&&a[i]<a[mique[mitail-1]]) mitail--;
    mique[mitail++]=i;
}

inline void mainque(int i)
{
    while(mahead<matail&&a[i]>a[maque[matail-1]]) matail--;
      maque[matail++]=i;
}

void outque(int pos)
{
    if(a[maque[mahead]]-a[mique[mihead]]>=k)
    {
        int nowlen=pos-now-1;
        int prelen=pre-now;
        ans+=(__int64)(nowlen+1)*nowlen/2;
        if(prelen>=1)
            ans-=(__int64)(prelen+1)*prelen/2;
        pre=pos-1;
    }
    while(a[maque[mahead]]-a[mique[mihead]]>=k)
        if(mique[mihead]<maque[mahead])
        {
            now=mique[mihead]+1;
            mihead++;
        }
        else
        {
            now=maque[mahead]+1;
            mahead++;
        }
}

int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        mihead=mahead=mitail=matail=0;
        pre=now=1;
        ans=0;
        for(i=1;i<=n;i++)
        {
            miinque(i);
            mainque(i);
            outque(i);
        }
        if(pre<n)
        {
           int nowlen=n-now;
           int prelen=pre-now;
           ans+=(__int64)(nowlen+1)*(nowlen)/2;
           if(prelen>=1)
            ans-=(__int64)(prelen+1)*(prelen)/2;
           pre=n-1;
        }

        printf("%I64d\n",ans+n);
    }
    return 0;
}

HDU 5089 Assignment（rmq+二分 或 单调队列）

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原文地址：http://blog.csdn.net/u014737310/article/details/46998227