DC poj 1064 Cable master

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题意是说有N条绳子,长度分别为Li,现在要从这N条绳子中产生K条长度相同的绳子,求这个最长长度。

算法是二分查找值算法,用一个函数C(x)判断是不是满足某种约定,然后根据判断结果不断更新这个区间

,最后得到这个最长长度,用二分法的话思想很简单,实现起来很容易。

 

 

#include<iostream>#include<algorithm>#define INF 9999999using namespace std;int N, M;int x[100001];//judge if it isbool C(int d){int pre = 0;for (int i = 1; i < M; i++){int cur = pre + 1;while (cur < N&&x[cur] - x[pre] < d){cur++;}if (cur == N){return false;}pre = cur;}return true;}int main(){while (cin >> N >> M){//inputfor (int i = 0; i < N; i++){cin >> x[i];}//sort the array xsort(x, x + N);int lb = 0, ub = INF;for (int i = 0; i < 100; i++){int mid = (lb + ub) / 2;if (C(mid)){lb = mid;}else{ub = mid;}}cout << lb << endl;}return 0;}


 

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