SGU101

题目大意

给定若干(n<=100)多米诺骨牌，每张骨牌两端有一数字(from 0 to 6)，求一种骨牌排列方式，使得除两端外相邻的两张骨牌所对的数字相同

如果将骨牌抽象成边，数字抽象成点，那么这个问题就变成了求一条有重边的欧拉路径，瞬间变得很裸，但前提是看的出

欧拉路径的求法，深搜即可

//Lib#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<ctime>#include<iostream>#include<algorithm>#include<vector>#include<string>#include<queue>using namespace std;//Macro#define rep(i,a,b) for(int i=a,tt=b;i<=tt;++i)#define rrep(i,a,b) for(int i=a,tt=b;i>=tt;--i)#define erep(i,e,x) for(int i=x;i;i=e[i].next)#define irep(i,x) for(__typedef(x.begin()) i=x.begin();i!=x.end();++i)#define read() (strtol(ipos,&ipos,10))#define sqr(x) ((x)*(x))#define pb push_back#define PS system("pause");typedef long long ll;typedef pair<int,int> pii;const int oo=~0U>>1;const double inf=1e20;const double eps=1e-6;string name="",in=".in",out=".out";//Varint n,map[10][10],d[10],cnt,st;struct T{int x,y;}c[108],s[108];void Init(){scanf("%d",&n);int a,b;rep(i,1,n){scanf("%d%d",&a,&b);map[a][b]++;map[b][a]++;d[a]++;d[b]++;c[i].x=a;c[i].y=b;}}void DFS(int u){rep(v,0,6)if(map[u][v]){--map[u][v];--map[v][u];DFS(v);cnt++;s[cnt].x=u;s[cnt].y=v;}}void Work(){rep(i,0,6)if(d[i]&1){cnt++;st=i;}else if(d[i]&&!st){st=i;}if(cnt>2){cout<<"No solution"<<endl;return;}cnt=0;DFS(st);if(cnt<n){cout<<"No solution"<<endl;return;}rrep(i,cnt,1)rep(j,1,n){if(s[i].x==c[j].x&&s[i].y==c[j].y){printf("%d +\n",j),c[j].x=-1;break;}else if(s[i].x==c[j].y&&s[i].y==c[j].x){printf("%d -\n",j),c[j].x=-1;break;}}}int main(){//freopen((name+in).c_str(),"r",stdin);//freopen((name+out).c_str(),"w",stdout);Init();Work();return 0;}