poj2456：Aggressive cows——贪心（二分+判定）

2022-02-08 08:32:22 阅读：135 来源： 互联网

标签：二分 distance poj2456 int pos param 判定 cows Aggressive

http://poj.org/problem?id=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?

最小值最大化、最大值最小化问题，解题方法为二分+判定函数

判定性问题：

判断distance是否可行
对distance进行二分

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXSIZE = 1E5 + 10;
int pos[MAXSIZE];   //隔间位置

/**
 * @brief 判定函数
 * 
 * @param n n个隔间位置
 * @param c c头牛
 * @param distance  两头牛之间的最小距离 
 * @return true distance<=任意两头牛之间距离
 * @return false 
 */
bool Judge(int n, int c, int distance){
    int current = pos[0];
    int number = 1;
    for(int i = 1; i < n; ++i){
        if(pos[i] - current >= distance){
            current = pos[i];
            number++;
        }
        if(number >= c){
            return true;
        }
    }
    return false;
}

int main(){
    int n, c;
    while(scanf("%d%d", &n, &c) != EOF){
        for(int i = 0; i < n; ++i){
            scanf("%d", &pos[i]);
        }
        sort(pos, pos + n);
        int left = 1;
        int right = pos[n-1] - pos[0];
        while(left <= right){
            int mid = left + (right - left) / 2;
            if(Judge(n, c, mid)){
                left = mid + 1;
            }
            else{
                right = mid - 1;
            }
        }
        printf("%d\n", right);
    }
    return 0;
}

标签：二分,distance,poj2456,int,pos,param,判定,cows,Aggressive
来源： https://www.cnblogs.com/dctwan/p/15870004.html

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poj2456：Aggressive cows——贪心（二分+判定）