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[50010],save[50010],L,N,M;

int main()
{
    while(~scanf("%d%d%d",&L,&N,&M))
    {
        int low = L,high = L;
        for(int i=0;i<N;i++)
        {
            scanf("%d",&dist[i]);
            low = min(low,dist[i]-dist[i-1]);
        }
        dist[N] = 0;
        dist[N+1] = L;
        N+=2;
        sort(dist,dist+N);

        int mid = (low+high)>>1;

        while(low <= high)
        {
            int rmv = 0,sum=0;
            for(int i=1;i<N;i++)
            {
                if(sum+dist[i]-dist[i-1] <= mid)
                {
                    rmv++;
                    sum+=dist[i]-dist[i-1];
                }
                else
                {
                    sum=0;
                }
            }

            if(rmv > M)
                high = mid-1;
            else
                low = mid+1;

            mid = (low+high)>>1;
        }
        printf("%d\n",low);
    }
}