标签：二分
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 to M 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
Output
Sample Input
25 5 2 2 14 11 21 17
Sample Output
4
题目大意：在n块石头中删除m块石头（不含起始，和结束的石头），求出题目中的最小距离的最大距离。
思路：二分问题。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
int a[50010];
int main()
{
int n,m,j,k,l;
while(~scanf("%d%d%d",&l,&n,&m))
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
a[0]=0;//将开始和结束的石头也加上。
a[n+1]=l;
int low=0,high=l;
int pos,sum;
int ans;
while(high>=low)
{
sum=0;
int mid=(high+low)>>1;
pos=0;
for(int i=1;i<=n+1;i++)
if(a[i]-a[pos]<mid)//比假设的最小值还小就当删除
sum++;
else
pos=i;
if(sum>m)
high=mid-1;
else
{
low=mid+1;
ans=mid;
}
}
printf("%d\n",ans);
}
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
POJ 3258 River Hopscotch(二分求最小中的最大)
标签：二分
原文地址：http://blog.csdn.net/grit_icpc/article/details/48023971
