poj1012 Joseph题解

DescriptionThe Joseph's problem is notoriously known. For those who are not familiar with the original problem: from among n people, numbered 1, 2, . . ., n, standing in circle every mth is going to be executed and only the life of the last remaining person will be saved. Joseph was smart enough to choose the position of the last remaining person, thus saving his life to give us the message about the incident. For example when n = 6 and m = 5 then the people will be executed in the order 5, 4, 6, 2, 3 and 1 will be saved. Suppose that there are k good guys and k bad guys. In the circle the first k are good guys and the last k bad guys. You have to determine such minimal m that all the bad guys will be executed before the first good guy. InputThe input file consists of separate lines containing k. The last line in the input file contains 0. You can suppose that 0 < k < 14. OutputThe output file will consist of separate lines containing m corresponding to k in the input file. Sample Input340Sample Output530

本题目采用链表来模拟这个过程，模拟的时候很简单，但是要注意以下问题，不然会超时的。

1.当某个人被out的时候，不能将链表中的节点delete掉，而是将所有的out的节点用一个指针保存起来，当下一个循环的时候，直接将其恢复就行，不然的话，删除了又要重新构建链表，肯定费时。

2.测试m是否成功时，不能简单的将m++，一个一个测试。我们注意到m只能为k+1的整数倍或者是k+1的整数倍加1，因为只剩下K+1个人时，cur索引必位于k或k+1上。这样m就可以一次自加(k+1)。

3.再有一点，程序上去后，由于测试数据k会重复出现，所以用一个数组result将已经算好的m值保存起来，以便下次直接采用。

#include<iostream>#include<vector>using namespace std;int result[13] = {0};struct Node{Node *prev;Node *next;bool saved;int attribute;};class LoopList{public:LoopList(int num = 0);bool Delete(int m);void Clear();void Reset();private:int size;int sizeOff;Node *head;Node *cur;Node *headOff;};LoopList::LoopList(int num /*= 0*/){headOff = NULL;sizeOff = 0;size = num;if(num==0){head = cur = NULL;}else{head = new Node;head->attribute = 1;//goodhead->saved = true;head->next = NULL;head->prev = NULL;cur = head;Node *tmp;for(int i=1;i<num;i++){tmp = new Node;if(i<num/2){tmp->attribute = 1;}else{tmp->attribute = 0;}tmp->saved = true;tmp->prev = cur;cur->next = tmp;cur = tmp;}cur->next = head;head->prev = cur;cur = head;}}bool LoopList::Delete(int m){m = m%size;if(m==0){cur = cur->prev;}else{while(--m){cur=cur->next;}}if(cur->attribute==1){return false;}else{sizeOff++;cur->prev->next = cur->next;cur->next->prev = cur->prev;Node *tmp = cur;cur = cur->next;if(headOff==NULL){headOff = tmp;tmp->next = tmp;tmp->prev = tmp;}else{headOff->prev->next = tmp;tmp->prev = headOff->prev;headOff->prev = tmp;tmp->next = headOff;}}size--;return true;}void LoopList::Clear(){cur = head->next;Node *tmp = cur;cur = cur->next;for(;cur!=head;cur=cur->next){delete tmp;tmp = cur;}cur = head;delete cur;size = 0;if(sizeOff==0){return;}tmp = cur = headOff;for(int i=1;i<sizeOff;i++){cur = cur->next;delete tmp;tmp = cur;}delete tmp;sizeOff = 0;return;}void LoopList::Reset(){size = size+sizeOff;sizeOff = 0;if(headOff!=NULL){cur = head->prev;headOff->prev->next = head;head->prev = headOff->prev;headOff->prev = cur;cur->next = headOff;headOff = NULL;}cur = head;}int SolveJoseph(int personNum){int m,i = 1;if(result[personNum-1]){m = result[personNum-1];return m;}int loopNum = 2*personNum;LoopList looplist(2*personNum);int flag = 0;while(1){if(flag==0){m = (personNum+1)*i;flag = 1;}else{m = (personNum+1)*i+1;i++;flag = 0;}int tmp = m%loopNum;if(tmp>=1 && tmp<=personNum){continue;}int out = 0;while(1){bool ok = looplist.Delete(m);if(ok){out++;}else{break;}if(out>=personNum){looplist.Clear();result[personNum-1] = m;return m;}}looplist.Reset();}looplist.Clear();return m;}int main(){int personNum;vector<int> vm;while(cin>>personNum && personNum>0){int m = SolveJoseph(personNum);vm.push_back(m);}for(int i=0;i<vm.size();i++){cout<<vm[i]<<endl;}return 0;}