[bzoj1650][Usaco2006 Dec][二分]跳石子

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.

数轴上有n个石子，第i个石头的坐标为Di，现在要从0跳到L，每次条都从一个石子跳到相邻的下一个石子。现在FJ允许你移走M个石子，问移走这M个石子后，相邻两个石子距离的最小值的最大值是多少。

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

5 rocks at distances 2, 11, 14, 17, and 21. Starting rock at position

0, finishing rock at position 25.

Sample Output

4

HINT

移除之前，最短距离在位置2的石头和起点之间；移除位置2和位置14两个石头后，最短距离变成17和21或21和25之间的4．

题解

二分最短距离的最大值，之后check
有几个细节，起点终点都有石头，排序后算差分的时候要减掉，不然会wa一些
还有，如果最后一段的和是小于二分出来的值的话，那么直接return false 因为最后一段无法再进行延伸，且全部取走的话方案不合法，那么return false

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>using namespace std;int L,n,m;int a[110000],b[110000];bool check(int p){    int cnt,ret,tmp;cnt=ret=tmp=0;//需要拿掉的石头数     int i=0;    while(i<=n)    {        while(cnt<p && i<=n){cnt+=b[++i];tmp++;}        if(i>n)tmp--;        if(cnt>=p){ret+=tmp-1;tmp=0;cnt=0;}    }    if(cnt<p && cnt!=0)return false;    if(ret<=m)return true;    return false;}int main(){    scanf("%d%d%d",&L,&n,&m);    for(int i=1;i<=n;i++)scanf("%d",&a[i]);    a[n+1]=L;    sort(a+1,a+1+n+1);    for(int i=1;i<=n;i++)b[i]=a[i]-a[i-1];    b[n+1]=a[n+1]-a[n];n++;    int l=1,r=L,ans;    while(l<=r)    {        int mid=(l+r)/2;        if(check(mid)){ans=mid;l=mid+1;}        else r=mid-1;    }    printf("%d\n",ans);    return 0;}