标签：
The Frog‘s Games
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3980 Accepted Submission(s): 1931
Problem Description
The annual Games in frogs‘ kingdom started again. The most famous game is the Ironfrog Triathlon. One test in the Ironfrog Triathlon is jumping. This project requires the frog athletes to jump over the river. The width of the river
is L (1<= L <= 1000000000). There are n (0<= n <= 500000) stones lined up in a straight line from one side to the other side of the river. The frogs can only jump through the river, but they can land on the stones. If they fall into the river, they
are out. The frogs was asked to jump at most m (1<= m <= n+1) times. Now the frogs want to know if they want to jump across the river, at least what ability should they have. (That is the frog‘s longest jump distance).
Input
The input contains several cases. The first line of each case contains three positive integer L, n, and m.
Then n lines follow. Each stands for the distance from the starting banks to the nth stone, two stone appear in one place is impossible.
Output
For each case, output a integer standing for the frog‘s ability at least they should have.
Sample Input
Sample Output
分析：先把石块sort一边，然后用二分枚举（0~L）青蛙跳跃距离，遍历出满足条件的最小值
代码示例：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int num[600000];
int main()
{
int L,n,m,i;
while(~scanf("%d%d%d",&L,&n,&m))
{
for(i=0;i<n;i++)
scanf("%d",&num[i]);
sort(num,num+n);
if(num[n-1]!=L)
num[n++]=L;
int l=0,r=L+1,mid,sum,h,flag;
while(l<r)
{
mid=(l+r)/2;
flag=0;
if(mid>=num[0]&&mid>=(num[n-1]-num[n-2]))
{
sum=1;
h=mid;
for(i=0;i<n;i++)
{
// if(mid==9)
// printf("%d\n",h);
if(h>=num[i]) continue;
else
{
sum++;
h=num[i-1]+mid;
if(sum>m||h<num[i])
break;
}
}
// printf("i=%d\n",i);
if(i>=n)
{
flag=1;
}
}
if(flag)
{
r=mid;
}
else
{
l=mid+1;
}
}
printf("%d\n",l);
}
return 0;
}
hdu 4004 The Frog's Games【二分】
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原文地址：http://blog.csdn.net/letterwuyu/article/details/42922385
