二分的几种简单应用——入门

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说起二分，最基础的二分搜索算法就不用了吧（最难的还际应用啊感觉）……实现起来很简单，搜索区间不断减半……唔……嘛……简单甩个模板好了(●‘?‘●)

下面开始学习二分的几种应用啦~~

1.假定一个解并判断是否可行

一般用于求解判断条件较为简单的最大化和最小化的问题，不断缩小答案的区间。

怎么说呢……举个栗子好了

Cable master POJ - 1064 

Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect computers for the contestants using a "star" topology - i.e. connect them all to a single central hub. To organize a truly honest contest, the Head of the Judging Committee has decreed to place all contestants evenly around the hub on an equal distance from it.
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible.
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter,and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled.
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.Input

Output

Sample Input

4 11
8.02
7.43
4.57
5.39

Sample Output

2.00

题意：给出n段绳子的长度，要求割出的段数，求每段的最大长度。
设bool型函数C（x）：每段绳长为x时，可以割出K段绳子
那么题目就变成了求x的最大值，一开始将答案的区间范围设为（0，inf）
我们所要做的就是不断缩小答案所在的区间，直至区间的左右边界最接近。
用二分的方法，在C（x（mid））返回true时，说明x偏小（x>=最大值），让x变大，取原区间的左区间中点继续判断；
                                                  返回false时，说明x偏大，让x变小，取原区间的左区间中点继续判断
直至区间缩小至最小，输出区间上界即可

不太明白为什么有的二分输出的是区间上界，有的是下界，有大佬给讲讲么

orz啊 ，WA了无数次……精度大坑啊……/(ㄒoㄒ)/~~记笔记记笔记
几个关于精度的问题：
①double转int的时候……记录每根绳子可以分为几段的时候crt=(int)(a[i]/x);加括号！！！不要忘了加括号！（敲黑板啊！）
②关于二分的循环次数，改用for(int i=0;i<100;i++)，避免eps选取不当陷入死循环的问题，100次循环可以达到1e-30的精度范围
③floor(<cmath>)向下取整
④double型scanf用%lf，printf用%f!!!!

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
double n,a[10005],sum;
int k;
bool C(double x)
{
    int crt=0;
    for (int i = 0; i < n; i++)
        crt +=(int)(a[i] / x);
    return crt >= k;
}
int main()
{
    while (scanf("%lf%d", &n,&k)!=EOF)
    {
        sum = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%lf",&a[i]);
        }
        double l = 0, h = 0x3f3f3f3f;
        for(int i=0;i<100;i++)
        {
            double mid = (l + h) / 2;
            if (C(mid)) l = mid;
            else h = mid;
        }
        printf("%.2f\n", floor(h * 100) / 100 );
    }
    return 0;
}

2.最大化最小值

道理跟上面差不多，最大化最小值或者最小化最大值的问题，通常可以用二分+贪心判断来解决

举个栗子，一起看一看

Aggressive cows POJ - 2456  

Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don‘t like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?Input

Output

Sample Input

5 3
1
2
8
4
9

Sample Output

3

Hint

#include<iostream>
#include<stdio.h>
using namespace std;
int n,c,a[100010];
bool C(int d)
{
    int t = a[0],crt=1;
    for (int i = 1; i <n; i++)
    {
        if (a[i] - t >= d)
        {
            t = a[i];
            crt++;
            if (crt >= c) return true;
        }
    }
    return false;
}
int main()
{
    while (scanf("%d%d", &n,&c)!=EOF)
    {
        for (int i = 0; i < n; i++)
            scanf("%d",&a[i]);
        sort(a, a + n);
        int l = 0, h = a[n-1]-a[0];
        while(l<h-1)
        {
            int mid = (l + h) / 2;
            if (C(mid)) l = mid;
            else h = mid;
        }
        printf("%d\n", l );
    }
    return 0;
}

3 2
1 1
1 2
1 3

1 2

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二分的几种简单应用——入门

标签：sub   else   表示   can   max   -o   other   mos   head   

原文地址：http://www.cnblogs.com/Egoist-/p/7401784.html