1650 跳石子 大视野评测

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．

 

Source

http://www.lydsy.com/JudgeOnline/problem.php?id=1650

 

这道题仍然用二分。 

因为包括端点，所以可以把n+2,存n+2个点，再排序。

注意从前到后check和从后到前check不一样，都要判断

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long long a[50001]={0},b[50001]={0};
long long l,m; 
int n;
bool check(long long k){
    for (int i=1;i<=n;i++) b[i]=a[i];
    int tot=0; bool f1=false,f2=false;
    for (int i=2;i<=n;i++){
        if (b[i]-b[i-1]<k) {
            tot++; 
            if (tot>m) { 
                f1=true;
                break;
            }
            b[i]=b[i-1];
        }
    } //从前到后判断
    for (int i=1;i<=n;i++) b[i]=a[n-i+1];
    for (int i=2;i<=n;i++){
        if (b[i]-b[i-1]<k) {
            tot++; 
            if (tot>m) { 
                f2=true;
                break;
            }
            b[i]=b[i-1];
        }
    } //从后到前判断
    if (f1&&f2) return false;
    return true;
}
void qsort(int left,int right){
    int i=left,j=right; long long mid=a[(left+right)/2];
    while (i<=j){
        while (a[i]<mid) i++;
        while (a[j]>mid) j--;
        if (i<=j){
            long long t=a[i]; a[i]=a[j]; a[j]=t;
            i++; j--;
        }
    }
    if (i<right) qsort(i,right);
    if (left<j) qsort(left,j);
}
int main(){
    scanf("%lld%d%lld",&l,&n,&m);
    a[1]=0;
    for (int i=2;i<=n+1;i++){        
        scanf("%lld",&a[i]);
    }
    qsort(1,n+1);
    a[n+2]=l; n++; n++;
    //for (int i=1;i<=n;i++) printf("%d ",a[i]);
    //printf("\n");
    long long left=0,r=l;
    long long ans=0;
    //printf("%d\n",check(0));
    while (left<=r) {
        long long mid=(left+r)/2; //printf("%d\n",mid);
        if (check(mid)) {
              left=mid+1;
              ans=mid;
        }
        else r=mid-1;
    }
    printf("%lld",ans);
}