二分法：青蛙游戏

The Frog's Games Time Limit:1000MS    Memory Limit:65768KB    64bit IO Format:%I64d & %I64u

Submit Status Practice ZOJ 3432

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
    

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

int l, m, n;
long long x, y, pla[500010];

bool ok(int mid){
    int placc = 0;
    int stp = 0;
    int pas = 0;
    while(placc < l)
    {
        placc += mid;         //跳过的总路程
        while(placc >= pla[pas+1])
            pas++;            //跳过的石头数
        placc = pla[pas];     //保证其落在石头上
        stp ++;               //跳跃次数
        //printf("d");
    }
    return stp > m;            //过不去
}

int main(){
    int mm = 0;
    int i, mid, ans, min;
    while(cin >> l >> n >> m){
        for(i = 1; i <= n; i ++){
            scanf("%I64d", &pla[i]);
        }
        sort(pla + 1, pla + n + 1);
        pla[0] = 0;
        pla[n + 1] = l;
        pla[n + 2] = 999999999;
        min = 0;
        for(int i=1; i<=n+1; i++)
           if(min < pla[i] - pla[i-1])
             min = pla[i] - pla[i-1];
        x = min;
        y = l;
        //printf("%d", min);
        while(y >= x){
            mid = (x + y)/2;
            if(ok(mid))          //过不去
                x = mid + 1;
            else{                //过得去
                y = mid - 1;
                ans = mid;
            }
        }
        cout << ans << endl;
    }
    return 0;
}