POJ 3258 River Hopscotch 二分-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_33414630/article/details/50768941

本文介绍了一种通过移除部分障碍来最大化跳跃距离的算法问题。该问题涉及在一条直线上分布的若干岩石间进行跳跃，目标是最小化连续岩石间的最大跳跃距离。采用二分搜索策略，最终确定在允许移除特定数量岩石的情况下，实现目标所需的最短跳跃距离。

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River Hopscotch

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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

Sample Output

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).

题目大意是一条长L 的河上，除了START 和 END 还有N 个石子，分别距离起点距离di，求去掉M个石子后相邻的最小距离的最大值。

一道比较典型的求最小值最大化的题目，所以大体思路是二分法。

注意要把START 和 END石头也存到数组。

判断时只要比较石头间距离和mid看是否需要去掉石头，再根据去掉的多少判断结果。（但是距离等于mid时也要去掉石头，不然会WA，有点不理解）

#include <iostream>
#include <algorithm>
#include<cstdio>
using namespace std;
bool cmp(int x,int y)
{
    return x<y;
}
int main()
{
    int l,n,m;
    while(cin>>l>>n>>m)
    {
        int len[50005],left,right,mid;
        len[0]=0;
        for(int i=1;i<=n;i++)
            cin>>len[i];
        len[n+1]=l;
        sort(len,len+n+2,cmp);
        left=0;
        right=l;
        while(left<=right)
        {
            int cnt=0,sum=0;
            mid=(left+right)/2;
            for(int i=1;i<=n+1;i++)
            {

                if((sum+=len[i]-len[i-1])<=mid)
                    cnt++;
                else
                    sum=0;
            }
            if(cnt<=m) left=mid+1;
            else right=mid-1;
        }
        cout<<left<<endl;
    }
}