PKU2456二分查找

原题http://poj.org/problem?id=2456

Aggressive cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6001		Accepted: 2989

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 3
1
2
8
4
9

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.
Source
USACO 2005 February Gold

//题目意思，求两头最近的牛之间的最大距离
//方法。直接对距离进行二分，找出答案

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
//#include <limlits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <vector>
#include <map>
using namespace std;
#define N 100005
int x
;
int n,c;

bool ok(int dis){
int i;
int next;
int last = 0;
for(i=1;i<c;i++){//要使所有的牛都要满足
next = last+1;
while(next<n && x[next]-x[last]<dis){//不断的扩大
next++;
}
if(next >= n){//因为此时牛的数量还没有满，说明了不能对所有的牛都满足，所以这个距离太大了
return false;
}
last = next;
}
return true;
}
int main(){
int i;

while(~scanf("%d%d",&n,&c)){
memset(x,0,sizeof(x));
for(i=0;i<n;i++){
scanf("%d",&x[i]);
}
sort(x,x+n);
int l = 0;
int r = 1000000000;//可能的距离
//int mid = l+(r-l)/2;
int mid;
while(r-l > 1){//不断的缩小距离，直到距离之间的距离为一
mid = (l+r)/2;
if(ok(mid) == true){//如果这个距离满足，说明了，还可以继续往大的来找
l = mid;
}
else{
r = mid;
}
}
printf("%d\n",l);
}

return 0;
}