101. Domino
time limit per test: 0.25 sec.
memory limit per test:
4096 KB
Dominoes – game played with small, rectangular blocks of wood
or other material, each identified by a number of dots, or pips, on its face.
The blocks usually are called bones, dominoes, or pieces and sometimes men,
stones, or even cards.
The face of each piece is divided, by a line or ridge,
into two squares, each of which is marked as would be a pair of dice...
The principle in nearly all modern dominoes games is to match one end of a piece to another that is identically or reciprocally numbered.
ENCYCLOP?DIA BRITANNICA
Given a set of domino pieces where each side is marked with two digits from 0 to 6. Your task is to arrange pieces in a line such way, that they touch through equal marked sides. It is possible to rotate pieces changing left and right side.
Input
The first line of the input contains a single integer N (1 ≤ N ≤ 100) representing the total number of pieces in the domino set. The following N lines describe pieces. Each piece is represented on a separate line in a form of two digits from 0 to 6 separated by a space.
Output
Write “No solution” if it is impossible to arrange them described way. If it is possible, write any of way. Pieces must be written in left-to-right order. Every of N lines must contains number of current domino piece and sign “+” or “-“ (first means that you not rotate that piece, and second if you rotate it).
Sample Input
5 1 2 2 4 2 4 6 4 2 1
Sample Output
2 - 5 + 1 + 3 + 4 -
题目链接：http://acm.sgu.ru/problem.php?contest=0&problem=101
把0-6当成点，输入的n个当成边。这样就形成了一个无向图。
答案就是求一个欧拉路径。
相关知识可以参考：http://blog.chinaunix.net/uid-26380419-id-3164913.html
一个是判断连通，然后度为奇数的点为0个或者2个，才有欧拉路径。
欧拉路径的求法dfs就可以了，很奇妙！
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2014-2-1 0:46:43 4 File Name :E:\2014ACM\SGU\SGU101.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 21 struct Edge 22 { 23 int to,next; 24 int index; 25 int dir; 26 bool flag; 27 }edge[220]; 28 int head[10],tot; 29 void init() 30 { 31 memset(head,-1,sizeof(head)); 32 tot = 0; 33 } 34 void addedge(int u,int v,int index) 35 { 36 edge[tot].to = v; 37 edge[tot].next = head[u]; 38 edge[tot].index = index; 39 edge[tot].dir = 0; 40 edge[tot].flag = false; 41 head[u] = tot++; 42 edge[tot].to = u; 43 edge[tot].next = head[v]; 44 edge[tot].index = index; 45 edge[tot].dir = 1; 46 edge[tot].flag = false; 47 head[v] = tot++; 48 } 49 int du[10]; 50 int F[10]; 51 int find(int x) 52 { 53 if(F[x] == -1)return x; 54 else return F[x] = find(F[x]); 55 } 56 void bing(int u,int v) 57 { 58 int t1 = find(u); 59 int t2 = find(v); 60 if(t1 != t2) 61 F[t1] = t2; 62 } 63 vector<int>ans; 64 void dfs(int u) 65 { 66 for(int i = head[u]; i != -1;i = edge[i].next) 67 if(!edge[i].flag ) 68 { 69 edge[i].flag = true; 70 edge[i^1].flag = true; 71 dfs(edge[i].to); 72 ans.push_back(i); 73 } 74 } 75 76 int main() 77 { 78 //freopen("in.txt","r",stdin); 79 //freopen("out.txt","w",stdout); 80 int n; 81 while(scanf("%d",&n) == 1) 82 { 83 init(); 84 int u,v; 85 memset(du,0,sizeof(du)); 86 memset(F,-1,sizeof(F)); 87 for(int i = 1;i <= n;i++) 88 { 89 scanf("%d%d",&u,&v); 90 addedge(u,v,i); 91 du[u]++; 92 du[v]++; 93 bing(u,v); 94 } 95 int s = -1; 96 int cnt = 0; 97 for(int i = 0;i <= 6;i++) 98 { 99 if(du[i]&1) cnt++; 100 if(du[i] > 0 && s == -1) 101 s = i; 102 } 103 bool ff = true; 104 if(cnt != 0 && cnt != 2) 105 { 106 printf("No solution\n"); 107 continue; 108 } 109 for(int i = 0; i <= 6;i++) 110 if(du[i] > 0 && find(i) != find(s)) 111 ff = false; 112 if(!ff) 113 { 114 printf("No solution\n"); 115 continue; 116 } 117 ans.clear(); 118 if(cnt == 0)dfs(s); 119 else 120 { 121 for(int i = 0;i <= 6;i++) 122 if(du[i] & 1) 123 { 124 dfs(i); 125 break; 126 } 127 } 128 for(int i = 0;i < ans.size();i++) 129 { 130 printf("%d ",edge[ans[i]].index); 131 if(edge[ans[i]].dir == 0)printf("-\n"); 132 else printf("+\n"); 133 } 134 } 135 return 0; 136 }
原文：http://www.cnblogs.com/kuangbin/p/3537491.html