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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

　

【題意】

有一條長度為L和河流，中間穿插n個石墩，青蛙跳m次經過石凳後到達對岸，求青蛙每次跳躍的最大距離的最小值

【分析】

1.二分左邊界left=1,右邊界right=L

2判斷mid=（left+right）/2是否可以在跳躍m次以內（包括m）到達對岸，具體做法通過貪心來完成，盡量每一次跳躍跨過的石凳最遠，這樣可以使跳躍的總次數最小

3不斷將區間二分，最終left==right時，left就是要求的值

【代碼】

#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;
} 

HDU 4004 The Frog's Games(二分答案)