POJ 3258(二分——最大化最小值)

River Hopscotch
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 14284 Accepted: 6059
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, L units 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 Di from the start (0 < Di < 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 2
2
14
11
21
17
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).
Source

USACO 2006 December Silver

题意：给N块石头和它们到岸边的距离，河长L，撤走M个石头，问需要跳跃最小值的最大值

做法：对这个最大值进行二分，也就是说设最大值为x,在这次过河过程中，跳跃的最小值是x能否过河，既然跳跃的最小值是x那么跳跃的长度不能低于x，即在当前位置di到di+x之间不能有石头，我们把这其中的石头都撤走，算出以这个最小值过河总共需要撤走多少石头并与m比较 ，注意要保证最后的位置在岸上，就是说最后一个石头到岸上的距离大于x才可以，我们可以把岸也当成一个石头在无穷远处，就可以保证一定最后在岸上了；

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define maxn 50010using namespace std;typedef long long ll;int l,m,n;int d[maxn];int q;bool judge(int x){    int ans=0;    int pos=0;    for(int i=0;i<n+2;i++)    {        if(d[i]-d[pos]>=x)        {            ans+=i-pos-1;            pos=i;        }    }    return ans<=m;}int main(){    while(scanf("%d%d%d",&l,&n,&m)==3)    {        for(int i=1;i<=n;i++)        {            scanf("%d",&d[i]);        }        d[0]=0;        d[n+1]=1e10;        int st=0,ed=l,mid=0,ans=0;        sort(d,d+n+2);        q=n*(n-1)/2;        while(ed>=st)        {            mid=(st+ed)/2;            if(judge(mid))            {                st=mid+1;                ans=mid;            }            else ed=mid-1;        }        printf("%d\n",ans);    }    return 0;}