Domino SGU

题意：n个诺米骨牌，每个骨牌的两面都有数字（0 ~ 6），求n个骨牌的一个排列，满足相邻骨牌相邻的一面的数字相同。

题解：每张骨牌当作一条边，啥意思呢？比如说第一张骨牌的两面是3和4，那么就在3和4之间连一条编号为1的无向边，建好图好，判断这张图能不能一笔画完，也就是一条欧拉路径能不能覆盖完所有的边。

PS：注意图可能是不连通，测试点4就是有多个连通块。

#include "stdafx.h"
#pragma warning(disable:4996)
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#define ll long long
#define P pair<int, int>
#define PP pair<int,pair<int, int>>
#define pb push_back
#define pp pop_back
#define lson root << 1
#define INF (int)2e9 + 7
#define rson root << 1 | 1
#define LINF (unsigned long long int)1e18
#define mem(arry, in) memset(arry, in, sizeof(arry))
using namespace std;

const int N = 100005;

int n, tot;
int head[10], use[300], deg[10];

vector<int> ans;

struct node { int to, next, id; } e[300];

void Inite() {
    tot = 0;
    mem(head, -1), mem(use, 0), mem(deg, 0);
}

void addedge(int u, int v, int id) {
    e[tot].to = v;
    e[tot].id = id;
    e[tot].next = head[u];
    head[u] = tot++;
}

void DFS(int u) {
    for (int i = head[u]; i != -1; i = e[i].next) {
        int v = e[i].to;
        if (!use[i]) {
            use[i] = use[i ^ 1] = 1;
            DFS(v);   
            ans.pb(e[i].id);
        }
    }
}

void Solve() {
    int cnt = 0, pos = -1;
    for (int i = 0; i <= 6; i++) if (deg[i] && (deg[i] & 1)) cnt++, pos = i;
    if (cnt != 0 && cnt != 2) puts("No solution");
    else {
        if (pos == -1) {
            for (int i = 0; i <= 6; i++) if (deg[i]) pos = i;
        }
        DFS(pos);
        if (ans.size() != n) {
            puts("No solution");
            return;
        }
        for(int i = ans.size() - 1; i >= 0; i--) {
            if (ans[i] > 0) printf("%d +
", ans[i]);
            else printf("%d -
", -ans[i]);
        }
    }
    ans.clear();
}

int main()
{
    while (scanf("%d",&n) != EOF) {
        Inite();
        for (int i = 1; i <= n; i++) {
            int u, v;
            scanf("%d %d", &u, &v);
            addedge(u, v, i);
            addedge(v, u, -i);
            deg[u]++;
            deg[v]++;
        }
        Solve();
    }
    return 0;
}