HDU 4004 The Frog's Games（青蛙过河，二分）

本文介绍了一项算法挑战——铁蛙三项中的跳跃比赛。任务是确定青蛙为成功跨越河流所需的最小跳跃能力，考虑河流宽度、石头分布及跳跃次数限制。

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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): 6136 Accepted Submission(s): 2985

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

  
   4
11

Source

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

题意：

一只小青蛙要过河，有 n 块石头，能蹦 m 次，求在过河后跳过最大距离的最小值。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define MYDD 11030

using namespace std;
typedef long long ll;

int dd[MYDD*100];
int L,n,m;

bool pan(int x) {
	int sum=0;//最好情况下跳的次数
	int last=0;//记录上一次的位置
	for(int j=1; j<=n+1;) {
		if(dd[j]>last+x) {
			if(last==dd[j-1]) {
				return false;
			}
			last=dd[j-1];
			sum++;
		} else
			j++;
	}
	sum++;
	return sum<=m;
}

int main() {
	while(scanf("%d%d%d",&L,&n,&m) != EOF) {
		for(int j = 1; j <= n; j++)
			scanf("%d",&dd[j]);
		sort(dd+1,dd+n+1);

		dd[n+1]=L;
		int turn=1,right=L;
		int ans;//记录最后的结果
		while(right>=turn) {
			int mid=(turn+right)/2;
			if(pan(mid)) {
				ans=mid;
				right=mid-1;
			} else {
				turn=mid+1;
			}
		}

		printf("%d\n",ans);
	}
	return 0;
}

超时代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define MYDD 11030

using namespace std;
typedef long long ll;

int dd[MYDD];
int L,n,m;

bool pan(int x) {
	int sum=0;//最好情况下跳的次数
	int last=0;//记录上一次的位置
	for(int j=1; j<=n+1;) {
		if(dd[j]>last+x) {
			if(last==dd[j-1]) {
				return false;
			}
			last=dd[j-1];
			sum++;
		} else
			j++;
	}
	sum++;
	return sum<=m;
} 
 
int main() {
	while(scanf("%d%d%d",&L,&n,&m) != EOF) {
		for(int j = 1; j <= n; j++)
			scanf("%d",&dd[j]);
		sort(dd+1,dd+n+1);

		dd[n+1]=L;
		int turn=1,right=L;
		int ans;//记录最后的结果
		while(right>=turn) {
			int mid=(turn+right)/2;
			if(pan(mid)) {
				ans=mid;
				right=mid-1;
			} else {
				turn=mid+1;
			}
		}
		
		printf("%d\n",ans);
	}
	return 0;
}