杭电4004 亚洲区预赛大连赛区-优快云博客

本文介绍了一道编程竞赛题目——铁蛙三项之跳河挑战。任务是确定青蛙为完成跳跃所需的最短能力距离。通过二分查找与贪心算法结合的方法解决此问题，并提供了一份AC代码示例。

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比赛时看了一下这道题，，感觉挺难的，后来一位学长想了出来。赛后听他讲了一下，对他是佩服不已，，，二分+贪心，，这个跨度很大啊，，，，，，，，，还是做的题太少了，比赛时完全不知道思路是什么，，纠结，，，，，题目：

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 
 

ac代码，其中用到了upper_bound函数

#include <iostream> #include <cstdio> #include <algorithm> using namespace std; int l,n,m; int a[500005]; bool judge(int x) { int pos=0,count=1; while(count<=m) { count+=1; //pos+=x; int num=upper_bound(a+1,a+n+2,pos+x)-(a+1); //printf("%d\n",num); pos=a[num]; if(pos==l) return true; } return false; } int main() { while(scanf("%d%d%d",&l,&n,&m)!=EOF) { for(int i=1;i<=n;++i) scanf("%d",&a[i]); a[n+1]=l; sort(a+1,a+n+2); int lleft=0,rright=l,mmax; while(lleft<=rright) { int mid=(lleft+rright)/2; //printf("%d %d %d \n",lleft,rright,mid); if(judge(mid)) { mmax=mid; rright=mid-1; } else lleft=mid+1; } printf("%d\n",mmax); } return 0; }