SGU101 Domino

题目大意

给出N张多米诺骨牌，骨牌两端各有一个数字(0 - 6)
现在要将这些骨牌排成一行，要求两张骨牌相邻一端的数字相同
构造一种骨牌的排列方案，其中骨牌可以翻转
无解则输出 “No solution”

算法思路

可以构建一张7个顶点的图，每张骨牌代表一条无向边
则符合要求的排列方案是该图的一个欧拉路径
可以通过DFS求解

时间复杂度: O(E)

代码

/** * Copyright (c) 2015 Authors. All rights reserved. *  * FileName: 101.cpp * Author: Beiyu Li <[e-mail]> * Date: 2015-05-21 */#include <bits/stdc++.h>using namespace std;#define rep(i,n) for (int i = 0; i < (n); ++i)#define For(i,s,t) for (int i = (s); i <= (t); ++i)#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)typedef long long LL;typedef pair<int, int> Pii;const int inf = 0x3f3f3f3f;const LL infLL = 0x3f3f3f3f3f3f3f3fLL;const int maxn = 7 + 5;const int maxe = 200 + 5;class Euler {        int n, psz;        struct Edge {                int v, w;                Edge *next;        } epool[maxe], *e[maxn], *cur[maxn];        int deg[maxn], vec[maxe], sz;        bool vis[maxe];        void dfs(int u)        {                while (cur[u]) {                        Edge *i = cur[u]; cur[u] = cur[u]->next;                        if (!vis[abs(i->w)]) {                                vis[abs(i->w)] = true;                                dfs(i->v);                                vec[sz++] = i->w;                        }                }        }        public:        void init(int n)        {                this->n = n; psz = 0;                memset(e, 0, sizeof(e));                memset(deg, 0, sizeof(deg));        }        void add_edge(int u, int v, int w)        {                Edge *i = epool + psz++;                i->v = v; i->w = w; i->next = e[u]; e[u] = i;                if (psz & 1) add_edge(v, u, -w);                else ++deg[u], ++deg[v];        }        bool find_path()        {                int s = -1, c = 0;                for (int i = 0; i < n; ++i) {                        if (deg[i] && s == -1) s = i;                        if (deg[i] & 1) s = i, ++c;                }                if (c && c != 2) return false;                sz = 0;                memcpy(cur, e, sizeof(cur));                memset(vis, false, sizeof(vis));                dfs(s);                if (sz < psz / 2) return false;                reverse(vec, vec + sz);                for (int i = 0; i < sz; ++i)                        printf("%d %c\n", abs(vec[i]), vec[i] > 0? '+': '-');                return true;        }} grp;int main(){        grp.init(7);        int m;        scanf("%d", &m);        For(i,1,m) {                int u, v;                scanf("%d%d", &u, &v);                grp.add_edge(u, v, i);        }        if (!grp.find_path()) puts("No solution");        return 0;}

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