Assignment HDU

题意：问有多少个连续的区间中的最大值和最小值的差不超过k。

做法，用RMQ（ST）预处理区间的最大值和最小值。然后枚举起点，二分求终点。时间复杂度为O（nlognlogn）

用cout还TLE了一发。。改了之后卡时过了（用log2慢些）

#include <iostream>#include <cstring>#include <algorithm>#include <cmath>#include <cstdio>#include <vector>#include <set>#include <map>#include <queue>using namespace std;int n,k;const int maxn = 200000+10;int a[maxn];int minn[maxn][40],maxx[maxn][40];void RMQ(){    memset(minn,0,sizeof(minn));    memset(maxx,0,sizeof(maxx));    for(int i=0;i<n;i++) {minn[i][0]=a[i];maxx[i][0]=a[i];}    for(int j=1;(1<<j)<n;j++)    {        for(int i=0;(i+(1<<j)-1)<n;i++)        {            minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);            maxx[i][j]=max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);        }    }    return ;}int ok(int bg,int mid){    int kk=0;    while((1<<(kk+1))<=mid-bg+1) kk++;//不用log2，比较慢    int ma=max(maxx[bg][kk],maxx[mid-(1<<kk)+1][kk]);    int mi=min(minn[bg][kk],minn[mid-(1<<kk)+1][kk]);    if(ma-mi<k) return 1;    else return 0;}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&k);        memset(a,0,sizeof(a));        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        RMQ();        long long ans=0;        for(int i=0;i<n;i++)        {            int l=i,r=n-1,res=0;//从            while(l<=r)            {                int mid=l+(r-l)/2;                if(ok(i,mid))                {                    res=mid;                    l=mid+1;                }                else r=mid-1;            }            ans+=(res-i+1);        }        printf("%I64d\n",ans);    }    return 0;}

用单调队列来做快多了。。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <cmath>#include <queue>#include <vector>using namespace std;const int maxn = 1e5+10;int a[maxn];int minn[maxn],maxx[maxn];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,k;        scanf("%d %d",&n,&k);        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        int front1=0,tail1=-1,front2=0,tail2=-1;        long long ans=0;        int j=0;        for(int i=0;i<n;i++)        {            while(front1<=tail1&&a[i]>a[maxx[tail1]]) tail1--;            maxx[++tail1] = i;            while(front2<=tail2&&a[i]<a[minn[tail2]]) tail2--;            minn[++tail2] = i;            while(a[maxx[front1]]-a[minn[front2]]>=k)            {                ans+=i-j;                if(a[j]==a[maxx[front1]]) front1++;                if(a[j]==a[minn[front2]]) front2++;                j++;            }        }        while(j<=n-1) {ans+=n-j;j++;}        printf("%I64d\n",ans);    }    return 0;}