[BZOJ1650]&&[Usaco2006 Dec]River Hopscotch 跳石子

描述 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。

输入格式 InputFormat

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.

输出格式 OutputFormat

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

样例输入 SampleInput

25 5 2

样例输出 SampleOutput


来源 Source


BZOJ 1650

代码 Code


#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf=0x7fffffff/27.11;
int a[50005];
int i,j,t,n,m,l,r,k,z,y,x,L,mid;
inline void read(int &x)
    x=0;char ch=getchar();
    while (ch<'0' || ch>'9') ch=getchar();
    while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
inline bool can()
    int i,j,t,l,r;
    while (r<=n)
        if (a[r]-a[l]>=mid)
    return t<=m;
int main()
    for (i=1;i<=n;i++) read(a[i]);
    while (l<=r)
        if (can()) l=mid+1;
        else r=mid-1;
    return 0;