sgu 158 分类： sgu 2015-05...

最新推荐文章于 2015-07-17 22:30:18 发布

转载最新推荐文章于 2015-07-17 22:30:18 发布 · 135 阅读

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原文链接：http://www.cnblogs.com/dashgua/p/4723027.html

本文深入解析了一种基于枚举的算法实现，通过具体实例详细介绍了其时间复杂度为O(L*(m+n))的计算过程。文章重点展示了如何利用C++进行编程实现，包括使用标准输入输出、数学运算、字符串处理等关键步骤。

枚举，时间复杂度 O(L∗(m+n))

#include<map>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<string>
#include<cstdio>
#include<utility>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>

const int MAXN = 305,MAXM = 305, INF = 1<<30, Nya = -1;

int L ,m ,n ;
int p[MAXM] = {0};
int d[MAXN] = {0};
int ansp ,ansl = Nya;

//dist(i, j, S) = |Dj + S - Pi|
int main()
{
#ifndef ONLINE_JUDGE
    freopen("sgu158.in","r",stdin);
    freopen("sgu158.out","w",stdout);
#endif

    std::cin >> L >> m;

    L <<= 1;

    for(int i = 1; i <= m ; i++)
     {std::cin >> p[i]; p[i]<<=1;}

    std::cin >> n; d[1] = 0; 
    for(int i = 2; i <= n ; i++)
     {std::cin >> d[i]; d[i]<<=1;}
    d[n+1] = INF;

    for(int s = 0; s + d[n] <= L; s++)
    {
        int cal = 0;
        for(int i = 1, j = 1; i <= m; i++)
        {
            while(d[j]+s <= p[i]) j++;
            cal += std::min(abs(d[j] + s - p[i]),abs(d[j-1] + s - p[i]));
        }
        if(cal > ansl) ansp = s, ansl = cal;
    }
    std::cout << (ansp>>1); 
    if(ansp&1) std::cout << ".5";
    std::cout << ' ';
    std::cout << (ansl>>1); 
    if(ansl&1) std::cout << ".5";

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;           
}