POJ3258-River Hopscotch-二分答案

一条河里有一串石头，给出石头间的间距，让你去掉m个石头，使最短间距最大。

二分答案，对于每一种mid，判断要不要删除这块石头。然后逼近答案。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int dist[],save[],L,N,M;

int main()
{
    while(~scanf("%d%d%d",&L,&N,&M))
    {
        int low = L,high = L;
        for(int i=;i<N;i++)
        {
            scanf("%d",&dist[i]);
            low = min(low,dist[i]-dist[i-]);
        }
        dist[N] = ;
        dist[N+] = L;
        N+=;
        sort(dist,dist+N);

        int mid = (low+high)>>;

        while(low <= high)
        {
            int rmv = ,sum=;
            for(int i=;i<N;i++)
            {
                if(sum+dist[i]-dist[i-] <= mid)
                {
                    rmv++;
                    sum+=dist[i]-dist[i-];
                }
                else
                {
                    sum=;
                }
            }

            if(rmv > M)
                high = mid-;
            else
                low = mid+;

            mid = (low+high)>>;
        }
        printf("%d\n",low);
    }
}