Marathon Ranking

题目描述：https://www.1point3acres.com/bbs/thread-225621-1-1.html

这道题是Bloomberg经典面试题，和LRU非常类似。维护一个list记录runner的顺序，一旦有runner经过sensor，我们就需要调整位置。本质其实就是要写一个 update(runner i, sensor j) 的函数。

注意C++中 list.splice 函数，由于对同一个list进行操作，指针不会失效，仍指向原来的list元素，runner2pos 也不用更新。

// https://www.1point3acres.com/bbs/thread-225621-1-1.htmlclass Marathon{int m, n; // m runner with id 1-m, n sensor with id 1-nlist<int> l; // record order of all runnersvector<list<int>::iterator> runner2pos; // runner_id -> the pos in listvector<int> passedSensor; // runner_id -> record how many sensors have been passedvector<int> firstPassed; // sensor_id -> the first runner who passed the sensorpublic:Marathon(int _m, int _n){m = _m;n = _n;runner2pos.resize(m+1);passedSensor.resize(m+1,0);firstPassed.resize(n+1,0);firstPassed[0] = 1; // at first in the order of 1~mfor (int i=1;i<=m;++i) l.push_back(i);for (auto it=l.begin();it!=l.end();++it){runner2pos[distance(l.begin(),it)+1]=it;// cout << distance(l.begin(),it)+1 << ' ' << *it << endl;
        }}void passMilestone(int runner_id){int sensor_id = ++passedSensor[runner_id];if (firstPassed[sensor_id]==0) firstPassed[sensor_id]=runner_id;// runner_id passed sensor_id, insert this runner to the front of previous sensorauto cur_pos=runner2pos[runner_id];auto new_pos=runner2pos[firstPassed[sensor_id-1]];l.splice(new_pos,l,cur_pos);}void printLeadBoard(){for (auto x:l) cout<<x<<' ';cout << endl;}
};int main() {Marathon m(3,2);m.passMilestone(3); // runner 3 passed sensor 1
    m.printLeadBoard();m.passMilestone(2); // runner 2 passed sensor 1
    m.printLeadBoard();m.passMilestone(2); // runner 2 passed sensor 2
    m.printLeadBoard();return 0;
}

转载于:https://www.cnblogs.com/hankunyan/p/11566707.html

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