POJ-3258 River Hopscotch 二分枚举求上限

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题目链接

题意：有一条河，河的长度已知，河中间有一些石头，石头的数量知道，相邻两块石头之间的距离已知。现在可以移除一些石头，问移除m块石头后，相邻两块石头之间的距离的最小值最大是多少。

思路：二分枚举答案。每次二分枚举一个值，判断该值能够去掉多少块石头。二分枚举求上限。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<math.h>#include<algorithm>#include<vector>#include<queue>using namespace std;const int maxn = 50005;const int inf = 1<<30;typedef __int64 LL;LL l;int n,m;LL rock[maxn];void fun(){LL ld = 0,rd = l,mid;int counts;while( ld < rd ){mid = (ld+rd)>>1;counts = 0;for( int i = 0,j = 1; j <= n+1;  )//找相邻点间最小距离{if( rock[j] - rock[i] <= mid ){j ++; counts ++;   //删除rock[j]}else{i = j;j ++;}}if( counts > m )rd = mid;elseld = mid+1;}printf("%d\n",rd);}int main(){//freopen("data.txt","r",stdin);while( scanf("%I64d%d%d",&l,&n,&m) != EOF ){for( int i = 1; i <= n; i ++ )scanf("%I64d",&rock[i]);rock[0] = 0;rock[n+1] = l;sort( rock,rock+n+1 );fun();}return 0;}

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