HDU 2993 MAX Average Problem (数形结合) #by Plato

http://acm.hdu.edu.cn/showproblem.php?pid=2993

题意：给你一段长度为n的数列, 求其长度不小于 K 的平均值最大的子串。

解析参考：http://blog.ac521.org/?p=565

大意是先将问题转化为求斜率的问题，然后将朴素的N^2的算法用下凸曲线维护（要推导一些性质），降为2N的复杂度

数形结合

#include <cstdio>
#include <iostream>
#include <fstream>
#include <cstring>
using namespace std;

int N,K;
int sum[100010];

struct point
{
    double x,y;
};

void read(int & a)
{
    char ch;
    while (ch = getchar(),ch < '0' || ch > '9');
    a = ch - '0';
    while (ch = getchar(),ch >= '0' && ch <= '9')
        a = a*10 + ch - '0';
}

double solve()
{
    double maxx = 0;
    static point s[100010];
    int head = 1,tail = 0;
    for (int j = K; j <= N; j++)
    {
        int k = j - K;
        while (tail-1 >= head &&  (s[tail].x - s[tail-1].x)*(sum[k] - s[tail].y) - (s[tail].y - s[tail-1].y)*(k - s[tail].x)<0)//if (tail-1 >= head)
        {
                tail --;
        }
        s[++tail].x = k;
        s[tail].y = sum[k];
        while (head+1 <= tail && (sum[j] - s[head].y)/(j - s[head].x) < (sum[j] - s[head+1].y)/(j - s[head+1].x))
            head++;
        double temp = (double)(sum[j] - s[head].y)/(j - s[head].x);
        //cout<<temp<<endl;
        if (temp > maxx) maxx = temp;
    }
    return maxx;
}

int main()
{
    freopen("test.txt","r",stdin);
    while(~scanf("%d%d",&N,&K))
    {
        sum[0] = 0;
        for (int i = 1; i <= N; i++)
        {
            read(sum[i]);
            sum[i] += sum[i-1];

        }

        printf("%.2lf\n",solve());
    }

    return 0;
}