Codeforces 508D Tanya and Password 欧拉通路Euler

来源：互联网发布：网件的访客网络编辑：程序博客网时间：2024/05/22 08:04

题目链接：点击打开链接

题意：给定n个长度为3各不相同的字符串，猜一个n+2位的密码。

这个密码包含上面的所有字符串

思路：

把 abc 拆成ab -> bc

则得到一个图，目标就是走遍所有 ab->bc 这样的边，也就是找一条欧拉通路。

#include <cstdio>#include <algorithm>#include <string.h>#include <queue>#include <cstring>#include <cmath>#include <iostream>#include <vector>using namespace std;typedef pair<int, int> pii;typedef long long ll;vector<pii>G[500000];int deg[4000];int ans[500000], top;int v[500000];int cur[4000];void eur(int s, int p){int f;while (cur[s]<G[s].size())if (v[G[s][f = cur[s]++].second] == 0){v[G[s][f].second] = 1;eur(G[s][f].first, G[s][f].second);} ans[top++] = p;}char c[200005][5];int n;int ch(char c){if ('0' <= c && c <= '9')return c - '0';if ('a' <= c && c <= 'z')return c - 'a' + 10;return c - 'A' + 36;}int getx(int id){return ch(c[id][0]) * 62 + ch(c[id][1]);}int gety(int id){return ch(c[id][1]) * 62 + ch(c[id][2]);}bool ok(){int x = 0, y = 0, pos = getx(0);for (int i = 0; i < 4000; i++){if (deg[i]>1 || deg[i] < -1)return false;if (deg[i] == 1) x++, pos = i;if (deg[i] == -1)y++;}if (x != y || x>1)return false;eur(pos, -1); top--;return top >= n;}void input(){memset(deg, 0, sizeof deg);memset(cur, 0, sizeof cur);memset(v, 0, sizeof v);scanf("%d", &n);for (int i = 0; i < n; i++){scanf("%s", c[i]);int x = getx(i), y = gety(i);G[x].push_back(pii(y, i));deg[x]++; deg[y]--;}}int main(){input();if (ok()){puts("YES");printf("%s", c[ans[top - 1]]);for (int i = top - 2; i>-1; i--)printf("%s", c[ans[i]] + 2);puts("");}else puts("NO");return 0;}

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