POJ 3258 River Hopscotch<二分>

River Hopscotch

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9326 Accepted: 4016

Description

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, Lunits away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance D_i from the start (0 < D_i < L).

To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.

Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N).

FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.

Input

Line 1: Three space-separated integers: L, N, and M 
Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

Output

Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

Sample Input

25 5 2214112117

Sample Output

4

Hint

Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

其实这几篇二分的题目，思路都一样，都是二分+判断函数，这个题和牛棚那个题都差不多，只不过这里是要去掉某个 石头，然后，第一个和最后一个不能去掉，所以我们可以转换一下，我们可以判断剩下的石头数是否为n-m+1（n,m同题目定义）,满足这个条件，说明二分传进来的值可行，我们继续像后寻找更大的，直到循环结束

#include<cstdio>#include<algorithm>using namespace std;int l,n,m;int a[50005];int f(int mid){    int cou=0,sum=0;    for(int i=0;i<n;i++)    {        if(a[i]-sum<mid)//如果小于，就舍去当前的石头            continue;        sum=a[i];        cou++;    }    if(l-a[n-1]>=mid)//判断最后一个石头能不能留下来        cou++;    if(cou>=n-m+1)        return 1;    else        return 0;}int main (){    while(~scanf("%d%d%d",&l,&n,&m)){        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        sort(a,a+n);        int ll=0,r=l,ans;        while(ll<=r){            int m=(ll+r)/2;            if(f(m)) {ll=m+1;ans=m;}//满足条件，继续往后寻找.            else {                r=m-1;            }        }        printf("%d\n",ans);    }}