The Frog's Games HDU

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 2225 3 311 218

Sample Output

411

每次做这种题都蒙圈，不过通过这几种的二分，自己终于抓住了这种问题 的本质，但愿以后不会卡这种题。

题意：青蛙跳河，河的宽度为L， 里面有n个石头，青蛙可以跳到石头上通过这些石头过河，青蛙最多可以跳m次，求青蛙最大弹跳能力的最小值。

思路： 贪心加二分，要求青蛙最大弹跳能力的最小值，那么我们就让青蛙尽量跳m次，并且每次都尽全力跳，

可以通过二分青蛙的弹跳能力来找到解.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int inf = 1000000100;int a[500100];int n,l,m;bool ok(int x){int fr = 0;int step = 0;for(int i=1; i<=n+1;){   if(fr >= l)break; // 一旦到了河对面就没必要再跳了，感觉这边也是个坑    int t = i;   while(a[i]-fr <= x && i<=n+1){t=i;++i;}       fr = a[t];   step++;   i = t;   if(step > m) return false;}    return true;}int main(){while(scanf("%d %d %d",&l, &n,&m) != EOF)    {a[0] = 0;for(int i=1; i<=n; i++){   scanf("%d", &a[i]);}a[n+1] = l;sort(a+1,a+1+n+1);        int ll = 0;int r =inf;int ans = 0;while(ll<=r){  int mid = (ll+r)/2;  if(ok(mid)){ans=mid; r=mid-1;}  else ll = mid+1;}          printf("%d\n",ans);}return 0;}

水波.