HDU 4004 The Frog's Games 二分

最新推荐文章于 2017-09-14 20:40:44 发布

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最新推荐文章于 2017-09-14 20:40:44 发布

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原文链接：http://www.cnblogs.com/Hyouka/p/5709520.html

本文介绍了一种解决铁蛙三项比赛中跳跃问题的算法。通过二分搜索确定青蛙需要具备的最小跳跃能力来成功越过河流。输入包含河流宽度、石头数量及最大跳跃次数等参数，输出青蛙至少需要达到的跳跃距离。

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The Frog's Games

Time Limit:1000MS Memory Limit:65768KB 64bit IO Format:%I64d & %I64u

Description

The annual Games in frogs' kingdom started again. The most famous game is the Ironfrog Triathlon. One test in the Ironfrog Triathlon is jumping. This project requires the frog athletes to jump over the river. The width of the river is L (1<= L <= 1000000000). There are n (0<= n <= 500000) stones lined up in a straight line from one side to the other side of the river. The frogs can only jump through the river, but they can land on the stones. If they fall into the river, they
are out. The frogs was asked to jump at most m (1<= m <= n+1) times. Now the frogs want to know if they want to jump across the river, at least what ability should they have. (That is the frog's longest jump distance).

Input

The input contains several cases. The first line of each case contains three positive integer L, n, and m.
Then n lines follow. Each stands for the distance from the starting banks to the nth stone, two stone appear in one place is impossible.

Output

For each case, output a integer standing for the frog's ability at least they should have.

Sample Input

6 1 2

25 3 3

Sample Output

二分青蛙能跳的距离

#include <iostream>
#include <cstdio>
#include <math.h>
#include <cstring>
#include <algorithm>
#include <iomanip>

using namespace std;

int a[500000];
int l,n,m;

bool llss(int s)
{
    int t,g;
    t=0;
    for(int i=1;i<=m;i++)
    {
        if(a[t+1]-a[t]>s)
            return false;
        g=t+1;
        while(a[g]-a[t]<=s)
        {
            if(g==n+1)
                return true;
            g++;
        }
        t=g-1;
    }
    return false;
}

int main()
{
    //freopen("input.txt","r",stdin);
    while(~scanf("%d%d%d",&l,&n,&m))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        a[0]=0;
        a[n+1]=l;
        sort(a,a+2+n);

        int min,max;
        min=1;
        max=a[n+1];
        int mid;
        while(min<max)
        {
            mid=(min+max)/2;
            if(llss(mid))
                max=mid;
            else
                min=mid+1;
        }
        printf("%d\n",max);
    }

}

转载于:https://www.cnblogs.com/Hyouka/p/5709520.html