首页 > 代码库 > [欧拉回路+手动开栈] poj 1780 Code
[欧拉回路+手动开栈] poj 1780 Code
题目链接：
http://poj.org/problem?id=1780
Code
Description KEY Inc., the leading company in security hardware, has developed a new kind of safe. To unlock it, you don‘t need a key but you are required to enter the correct n-digit code on a keypad (as if this were something new!). There are several models available, from toy safes for children (with a 2-digit code) to the military version (with a 6-digit code). The safe will open as soon as the last digit of the correct code is entered. There is no "enter" key. When you enter more than n digits, only the last n digits are significant. For example (in the 4-digit version), if the correct code is 4567, and you plan to enter the digit sequence 1234567890, the door will open as soon as you press the 7 key. The software to create this effect is rather simple. In the n-digit version the safe is always in one of 10n-1 internal states. The current state of the safe simply represents the last n-1 digits that have been entered. One of these states (in the example above, state 456) is marked as the unlocked state. If the safe is in the unlocked state and then the right key (in the example above, 7) is pressed, the door opens. Otherwise the safe shifts to the corresponding new state. For example, if the safe is in state 456 and then you press 8, the safe goes into state 568. A trivial strategy to open the safe is to enter all possible codes one after the other. In the worst case, however, this will require n * 10n keystrokes. By choosing a good digit sequence it is possible to open the safe in at most 10n + n - 1 keystrokes. All you have to do is to find a digit sequence that contains all n-digit sequences exactly once. KEY Inc. claims that for the military version (n=6) the fastest computers available today would need billions of years to find such a sequence - but apparently they don‘t know what some programmers are capable of... Input The input contains several test cases. Every test case is specified by an integer n. You may assume that 1<=n<=6. The last test case is followed by a zero. Output For each test case specified by n output a line containing a sequence of 10n + n - 1 digits that contains each n-digit sequence exactly once. Sample Input 1 2 0 Sample Output 0123456789 00102030405060708091121314151617181922324252627282933435363738394454647484955657585966768697787988990 Source Ulm Local 2004 |
[Submit] [Go Back] [Status] [Discuss]
题目意思：n位密码，每位可以是0~9任意一个，求一个串，使得这个串中任意两个n位都不相同。
解题思路：
欧拉路径+手动开栈
欧拉路径求法是dfs走到不能走的时候后纪录路径，然后逆序输出。
代码：
//#include<CSpreadSheet.h>
#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
//#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 1000000
int n,s;
int List[Maxn+10],St[Maxn+10];
char ans[Maxn+10];
void solve(int cur,int M)
{
while(List[cur]<10)
{
int w=cur*10+List[cur];
List[cur]++;
St[++s]=w;
cur=w%M;
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(~scanf("%d",&n)&&n)
{
int M=pow(10.0,double(n-1));
for(int i=0;i<M;i++)
List[i]=0;
s=0;
int a=0;
solve(0,M);
while(s)
{
ans[++a]=St[s]%10+'0';
s--;
solve(St[s+1]/10,M);
}
for(int i=1;i<n;i++)
printf("0");
while(a)
printf("%c",ans[a--]);
putchar('\n');
}
return 0;
}
[欧拉回路+手动开栈] poj 1780 Code
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。