POJ-2456 Aggressive cows

Aggressive cows

Time Limit: 1000MS Memory Limit: 65536K

Description

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

* Line 1: Two space-separated integers: N and C 

* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

* Line 1: One integer: the largest minimum distance

Sample Input

5 312849

Sample Output

3

Hint

OUTPUT DETAILS: 

FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3. 

Huge input data,scanf is recommended.

————————————————————二分的分割线————————————————————

思路：二分逼近的模板题。二分法是一种思想。对答案进行二分的时候，意味着你在解决问题之前，事先知道了答案的区间。或者比答案大一些的区间。二分查找模板如下：

int binSearch(int a[], int l, int r, int key){    int mid;    while(l <= r){        mid = (l+r) >> 1;        if(a[mid] == key)            return mid;        else if(a[mid] < key)            l = mid + 1;        else            r = mid - 1;    }    return -1;}

要灵活运用二分的思想，不只是二分查找的模板所能解决的。本题要求“最大的最小值”，即在可行范围内尽量使牛之间的距离远，最小的答案是1,最大的答案绝不会超过Max(a[i])。考虑Check函数怎么写。既然二分的是答案（距离），那么check的时候枚举下标。一个接一个考查 当前cow下标 + m > 下一cow下标？ 超过的话，该下标不能放牛。能放的话，cnt++，更新当前cow下标。

代码如下：

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>void scanf_(int &num){ //输入外挂    char in;    bool neg = false;    while(((in = getchar()) > '9'||in < '0')&&in != '-') ;    if(in == '-') {        neg = true;        while((in = getchar()) >'9'||in < '0') ;    }    num = in - '0';    while(in = getchar(), in >= '0'&&in <= '9')  num *= 10, num += in - '0';    if(neg) num = 0 - num;}int a[100010];int n, c;int cmp(const void* a, const void* b) {    return *(int *)a - *(int *)b;}bool check(int m) {    int cnt = 1;//数牛    int last = a[0];//第一头牛    for(int i = 1; i < n; i++) {        if(last + m > a[i]) continue;//找到满足该解的a[i]        else {            last = a[i];//从a[i]开始看下一头牛            cnt++;//又放了一头牛            if(cnt == c)    return true;//直到牛全部放完或者放不完        }    }    return false;}int main() {    int l = 1, r = -1;    scanf("%d%d", &n, &c);    for(int i = 0; i < n; i++) {        scanf_(a[i]);        if(r < a[i])    r = a[i];    }    qsort(a, n, sizeof(a[0]), cmp); //建立“下”标轴    int mid;    int ans = -1;    while(l <= r) {        mid = (l+r) >> 1;        if(check(mid))  l = mid + 1;        else            r = mid - 1;    }    printf("%d\n", r);return 0;}