hdu4004之二分查找

The Frog's Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2554    Accepted Submission(s): 1315

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

6 1 2225 3 311 218

 

Sample Output

411

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX=500000+10;int s[MAX];int search(int left,int right,int v){//查找最右端<=v的数 while(left+1<right){//剩下两个元素的时候另外判段,如果left<right可能会无限循环 int mid=left+right>>1;if(s[mid]<=v)left=mid;else right=mid-1;}if(s[right]<=v)return right;return left;}int judge(int v,int n,int m){int num=0,left=0;while(left<n && ++num<=m)left=search(left,n,s[left]+v);//进行二分查找并且记录查找次数(即跳跃次数) return num;}int main(){int L,n,m;while(scanf("%d%d%d",&L,&n,&m)!=EOF){for(int i=1;i<=n;++i)scanf("%d",s+i);sort(s,s+n+1);s[n+1]=L;int left=0,right=L,mid;while(left<right){//对于青蛙可能跳的距离进行二分 mid=left+right>>1;if(judge(mid,n+1,m)<=m)right=mid;//判断青蛙以跳跃距离mid最少跳几次到L else left=mid+1;}cout<<right<<endl;}return 0;}