HDU-4004-The Frog's Games（二分）

港控/mmm° 2022-05-28 03:38 131阅读 0赞

## The Frog's Games ##

### 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

25 3 3

### Sample Output ###

#include<bits/stdc++.h>
    using namespace std;
    int t,n,m;
    int a[500006];
    int judge(int x)
    {
        int sum=0;
        int k=0;///原点
        for(int i=1;i<=n;i++)
        {
            if(a[i]-a[k]<=x&&a[i+1]-a[k]>x)
            {
                sum++;
                k=i;///更新原点
            }
        }
        sum++;
        if(sum<=m)
        {
            return 1;
        }
        else
        {
            return 0;
        }
    }
    int main()
    {
        while(~scanf("%d%d%d",&t,&n,&m))
        {
            a[0]=0;
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
            }
            a[n+1]=t;
            sort(a,a+n+2);
            int Max=0;
            for(int i=0;i<=n;i++)
            {
                if(Max<a[i+1]-a[i])
                {
                    Max=a[i+1]-a[i];///如果小于这个值，那就一定跳不过去，就不用枚举了。
                }
            }
            int l,r,mid,flag;
            l=Max;
            r=t;
            mid=(r+l)/2;
            for(;;)
            {
                if(l>r)
                    break;
                if(judge(mid))///如果这个能跳过
                {
                    flag=mid;
                    r=mid-1;///找更短的获取更优
                    mid=(r+l)/2;
                }
                else
                {
                    l=mid+1;
                    mid=(r+l)/2;
                }
            }
            printf("%d\n",flag);
        }
        return 0;
    }

二分枚举找最优。从能跳的max到一次跳到终点的最大能力之间找到一个能力点，并试着找到更优的解法，最终便是结果