POJ 3258-River Hopscotch（二分法-最大化最短距离）

River Hopscotch

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12505		Accepted: 5358

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 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 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

题目意思：

从起点到终点，即0到L这段路程中有N块石头，奶牛每次只能从前一块跳到后一块石头上去。

想要移除除了起点终点之外的M块石头，使得奶牛从起点到终点跳跃的最短距离最大。

解题思路：

添加起终点的石头0和L。

对移除的石头个数二分，low是奶牛从起点到终点跳跃的最短距离，high是整个路程L。

①枚举各个石头，计算其与前一块石头之间的距离，然后相加起来：

如果比最短距离小，说明距离短了，可以移除当前石头来增大距离，移除的石头数量加1；

否则说明距离够大，置零后重新计算下一段距离。

②枚举完毕后：

如果移除的石头数量大于M，说明距离偏大，范围变成前二分之一；

否则说明距离偏小，范围变成后二分之一。

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
#define INF 0xfffffff
#define MAXN 50050
int a[MAXN];
int l,n,m,high,low,mid;
bool solve()
{
    int res=0,sum=0;
    for(int i=1; i<=n+1; ++i)
    {
        int t=a[i]-a[i-1];//与前一块石头之间的距离
        sum+=t;//若干块距离之和
        if(sum<=mid) ++res;
        else sum=0;
    }
    if(res<=m) return true;
    else return false;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>l>>n>>m;
    high=l,low=l;
    a[0]=0;
    a[n+1]=l;
    for(int i=0; i<=n; ++i)
        cin>>a[i];
    sort(a,a+n+2);
    for(int i=1; i<=n+1; ++i)
    {
        int t=a[i]-a[i-1];//与前一块石头之间的距离
        if(t<low) low=t;
    }
    while(low<=high)
    {
        mid=0.5*(low+high);
        if(solve()) low=mid+1;
        else high=mid-1;
    }
    cout<<low<<endl;
    return 0;
}
/*
25 5 2
2
14
11
21
17
*/