The Frog's Games（二分）

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原文链接：http://www.cnblogs.com/handsomecui/p/5523042.html

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#java #数据结构与算法

本文介绍了一项关于铁蛙三项中跳远比赛的算法挑战。比赛中青蛙需跨越宽度为L的河流，借助河中n块石头最多跳跃m次完成任务。通过二分查找确定青蛙至少需要具备的跳跃能力，即最长跳跃距离。文章提供了完整的C++实现代码。

The Frog's Games

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 5676 Accepted Submission(s): 2732

Problem 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 2 25 3 3 11 2 18

Sample Output

4 11

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

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lcy

题解：简单二分：

代码：

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<set>
#include<vector>
using namespace std;
const int MAXN = 500010;
int L, n, m;
int ston[MAXN];
bool js(int x){
    int cnt = 0, last = ston[0];
    for(int i = 1; i <= n; i++){
        if(ston[i] - ston[i - 1] > x)
            return false;
        if(ston[i] - last > x){
            cnt++;
            last = ston[i - 1];
            if(cnt >= m)
                return false;
        }
    }
    return true;
}
int erfen(int l, int r){
    int mid, ans;
    while(l <= r){
        mid = (l + r) >> 1;
        if(js(mid)){
            ans = mid;
            r = mid - 1;
        }
        else
            l = mid + 1;
    }
    return ans;
}
int main(){
    while(~scanf("%d%d%d", &L, &n, &m)){
        for(int i = 1; i <= n; i++)
            scanf("%d", ston + i);
        ston[0] = 0;
        ston[n + 1] = L;
        n++;
        sort(ston, ston + n + 1);
        printf("%d\n", erfen(0, L));
    }
    return 0;
}