HDU 4004 The Frog's Games(二分答案)

The Frog's Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 10307    Accepted Submission(s): 4686

　

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

 

　

Recommend

lcy

　

【题意】

有一条长度为L和河流，中间穿插n个石墩，青蛙跳m次经过石凳后到达对岸，求青蛙每次跳跃的最大距离的最小值

【分析】

1.二分左边界left=1,右边界right=L

2判断mid=（left+right）/2是否可以在跳跃m次以内（包括m）到达对岸，具体做法通过贪心来完成，尽量每一次跳跃跨过的石凳最远，这样可以使跳跃的总次数最小

3不断将区间二分，最终left==right时，left就是要求的值

【代码】

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=5e5+5;
int n,m,len,dis[N];
bool judge(int now){
	if(dis[1]>now) return 0;
	int sp=0,tot=0;
	for(int i=1;i<=n;){
		if(dis[i]-sp<=now){
			if(dis[i]-sp==now||i==n){
				tot++;
				sp=dis[i];
			}
			i++;
		}
		else{
			tot++;
			sp=dis[i-1];
			if(dis[i]-sp>now) return 0;
		}
	}
	return tot<=m;
}
int main(){
	while(scanf("%d%d%d",&len,&n,&m)==3){
		for(int i=1;i<=n;i++) scanf("%d",dis+i);dis[++n]=len;
		sort(dis+1,dis+n+1);
		int l=1,r=len,mid,ans=0;
		while(l<=r){
			int mid=l+r>>1;
			if(judge(mid)){
				ans=mid;
				r=mid-1;
			}
			else l=mid+1;
		}
		printf("%d\n",ans);
	}
	return 0;
}