HDU 4004 二分

Posted 2020-09-26 zzq

The Frog‘s Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 7307    Accepted Submission(s): 3492

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 2 2 25 3 3 11 2 18

 

 

Sample Output

4 11

 

 

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 久违的惨案，错误tm竟然是因为没排序就去找最大差值只能怪自己傻逼了，很水的一个二分调到结束=  =

#include<bits/stdc++.h>
using namespace std;
int a[500005]={0},L,M,maxn,N;
int solve(int k)
{
   if(maxn>k) return 0;
   int s=1,x=0,i,j;
   for(i=1;i<=N;++i)
       if(a[i]-a[x]<=k&&a[i+1]-a[x]>k) {x=i;s++;}
   return s<=M;
}
int main()
{

    int i,j,k;
    while(cin>>L>>N>>M){maxn=0;
        for(i=1;i<=N;++i) {
                scanf("%d",&a[i]);
        }a[N+1]=L;
        sort(a+1,a+1+N);   //由于输入时的无序所以要排序
        for(i=1;i<=N+1;++i) maxn=max(maxn,a[i]-a[i-1]);   //第一次在输入时做的这一步导致整个的失败哎，以后要注意！
        int l=0,r=L,mid;
        while(l<r){
            mid=l+(r-l)/2;
            if(solve(mid)){
              r=mid;
            }
            else{
              l=mid+1;
            }
        }
        cout<<l<<endl;
    }
    return 0;

}