POJ - 1780 Code （歐拉回路+手寫DFS）

POJ - 1780 Code （歐拉回路+手寫DFS）

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 10^n-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

題意：破解一套1-6位長度密碼的系統，尋找這樣一個序列：對於N位的密碼10^N+N-1長度的連續的長為N的串能夠枚舉完所有的密碼。

思路：做的頭疼，懶得寫了。就拿別人的題解吧：對於每一個長度為n的串，讓該串的前n-1位為一個節點，後n-1位為另一個節點這樣就確定了這個串。

n 位數有10n 種編碼方案（即10n 組數），要使得一個數字序列包含這10n 組n 位數，且序列的長

度最短，唯一的可能是每組數出現一次且僅一次、且前一組數的後n-1 位是後一組數的前n-1 位，

這樣10n 組數各取1 位，共10n 位，再加上最後一組數的後n-1 位，總位數是10n + n - 1。

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1000015;

struct Edge {
	int v, next;
} e[maxn];
int n, mod, head[maxn], stk[maxn];
int idx, lim, edge, top;
char vis[maxn];
 
void insert(int a, int b) {
	e[idx].v = b;
	e[idx].next = head[a];
	head[a] = idx++;
}

void dfs() {
	stk[top++] = 0;
	while (1) {
		int flag = 0;
		int v = stk[top-1];
		if (top >= idx-(n-2))
			return;
		for (int i = head[v]; i != -1; i = e[i].next)
			if (!vis[i]) {
				flag = 1;
				vis[i] = 1;
				stk[top++] = e[i].v;
				break;
			}
		if (!flag)
			--top;
	}
}

int main() {
	while (scanf("%d", &n) != EOF && n) {
		if (n == 1) {
			printf("0123456789\n");
			continue;
		}
		idx = top = 0;
		memset(head, -1, sizeof(head));
		memset(vis, 0, sizeof(vis));
		lim = 1;
		for (int i = 1; i < n; i++)
			lim *= 10;
		edge = lim * 10;
		mod = lim / 10;
		for (int i = 0; i < lim; i++)
			for (int j = 9; j >= 0; j--)
				insert(i, (i%mod)*10+j);
		dfs();
		for (int i = 0; i < top-1; i++)
			printf("%d", stk[i]/mod);
		printf("%d", stk[top-1]);
		for (int i = 1; i < n; i++)
			printf("0");
		printf("\n");
	}
	return 0;
}