HDU 5289 Assignment

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题意：在n个数中找一共有几个数组，使得数组内的最值差不超过k，数组元素要求连续

解一：ST

#include <string.h>
#include <math.h>
#include <stdio.h>
#define ll __int64
#define MAX(a,b) ((a)<(b)?(b):(a))
#define MIN(a,b) ((a)<(b)?(a):(b))
const int maxn=100005;
ll a[maxn];
int maxx[maxn][20];
int minn[maxn][20];
void rmq(int n)           //子串赋值
{
    int i,j;
    for(i=1;(1<<i)<=n;i++)
        for(j=1;(j+(1<<i)-1)<=n;j++)
        {
            int temp=1<<(i-1);
            maxx[j][i]=MAX(maxx[j][i-1],maxx[j+temp][i-1]);
            minn[j][i]=MIN(minn[j][i-1],minn[j+temp][i-1]);
        }
    return ;
}
int ask(int l,int r)
{
    int temp=int(log(r-l+1)/log(2));
    int ma=MAX(maxx[l][temp],maxx[r-(1<<temp)+1][temp]);
    int mi=MIN(minn[l][temp],minn[r-(1<<temp)+1][temp]);
    return ma-mi;
}
int main()
{
    int t,n,k;
    ll sum;
    //freopen("1002.in","r",stdin);
    //freopen("out.out","w",stdout);
    while(scanf("%d",&t)!=-1)
    {
        while(t--)
        {
            sum=0;
            scanf("%d%d",&n,&k);
            for(int i=1;i<=n;i++)
            {
                scanf("%I64d",&a[i]);
                minn[i][0]=maxx[i][0]=a[i];
            }
            rmq(n);
            for(int i=1;i<=n;i++)
            {
                int fr=i,ed=n;
                while(fr<ed)              //二分查找
                {
                    int temp=(fr+ed)/2;
                    if(ask(i,temp)>=k)
                        ed=temp-1;
                    else
                        fr=temp+1;
                }
                if(ask(i,ed)<k)           //坐标差值就是从i点为起点一共有多少个满足的子串
                    sum+=((ll)(ed-i+1));
                else
                    sum+=((ll)(ed-i));
            }
            printf("%I64d\n",sum);
        }
    }
    return 0;
}

HDU 5289 Assignment

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原文地址：http://blog.csdn.net/trq1995/article/details/47003629