H - River Hopscotch 문제 풀이 보고서(장호성륜)

H - River Hopscotch
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
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Description
Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L).
To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.
Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up toM rocks (0 ≤ M ≤ N).
FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.
Input
Line 1: Three space-separated integers: 
L, 
N, and 
M 
Lines 2..
N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.
Output
Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing 
M rocks
Sample Input

25 5 2
2
14
11
21
17

Sample Output

4

Hint
Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).
제목 링크:http://poj.org/problem?id=3258
해법 유형: 2분
문제 풀이 사고방식: 이 문제는 나로 하여금 이분법에 대해 새로운 인식을 가지게 했다.그 중의 함수 관계를 주의해라. 두 돌 사이의 최소 거리와 이동하는 돌 수 사이는 단조로운 점증 함수이기 때문에 거리를 2분할 수 있고 효율이 매우 높다.
알고리즘 구현:

//STATUS:C++_AC_125MS_316K
#include<stdio.h>
#include<stdlib.h>
const int MAX=50010;
int cmp_num(const void *a,const void *b);
int d[MAX];
int main()
{
//	freopen("in.txt","r",stdin);
	int L,N,M,i,low,mid,high,tot,a,b;
	while(~scanf("%d%d%d",&L,&N,&M))
	{
        low=-1,high=L+1,mid=L;
		for(i=0;i<N;i++)
			scanf("%d",&d[i]);
		
		qsort(d,N,sizeof(int),cmp_num);   // 

		if(N && N>M)   // N=0 N<=M 
			for(mid=(low+high)/2;high-low!=1;mid=(low+high)/2){
				for(i=0,a=0,b=d[0],tot=0;i<N;a=b,b=d[++i]){
					while(b-a<mid && i<N){       // ， mid；
						tot++;
						b=d[++i];
					}
					if(i==N-1){if(L-d[N-1]<mid)tot++; break;}   // 
					if(i==N){if(L-a<mid)tot++; break;}
				}
       
	           	if(tot<=M)low=mid;
	        	else high=mid;
			}
		printf("%d
",mid);
	}
	return 0;
}

int cmp_num(const void *a,const void *b)   // 
{
	return *(int*)a - *(int*)b;
}