sgu101 欧拉路

题意：给你n张牌，每个牌有两面，要求找出一种摆放顺序，使得相邻的牌的面相同。初始给出的牌的方向为正

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <stack>

#include <queue>

#include <vector>

#include <map>

#include <string>

#include <iostream>

using namespace std;

int father[7];

int len;

int head[7];

int bx;

int degree[7];

int vis[7];

int getfather(int x)

{

    if(father[x]!=x) father[x]=getfather(father[x]);

    return father[x];

}



struct edge

{

    int val;int to;int next;int vis;int oppo;int t;

}e[1111];



void link(int a,int b)

{

    int fa=getfather(a); int fb= getfather(b);

    father[fa]=fb;

}

void add(int from,int to,int val)

{

    e[len].t=1;

    e[len].to =to ;e[len].val = val;e[len].vis=0;

    e[len].oppo= len+1;

    e[len].next= head[from];



    head[from]= len++;

    e[len].to= from;e[len].val=val;

    e[len].t= 0;

    e[len].vis=0;

    e[len].oppo= len-1;

    e[len].next= head[to];

    head[to]=len++;

}



int judge()

{

    bx= 0 ;

    int ans=0;int ans1=0;

    for(int i=0;i<=6;i++)

    if(vis[i])if(father[i]==i) ans++;

    if(ans!=1) return 0;

    for(int i=0;i<=6;i++)

        if(degree[i]&1) ans1++,bx= i;

    if(bx==0) for(int i=0;i<=6;i++)

        if(vis[i]) bx= i;

    if(ans1==2||ans1==0) return 1;

    return 0;

}



struct Node

{

    int t;int x;

};

stack <Node> q;

void dfs(int x)

{

    for(int i= head[x];i!=-1;i=e[i].next){

        if(e[i].vis) continue;

        int cc=e[i].to;

        e[i].vis= 1;e[e[i].oppo].vis=1;

        dfs(cc);

        Node gg ;

        if(e[i].t==1) gg.t=1;

        else gg.t=0;

        gg.x= e[i].val;

        q.push(gg);

    }

}

int main()

{

    int n,a,b;

    scanf("%d",&n);

        memset(head,-1,sizeof(head));

        len=0;

        memset(vis,0,sizeof(vis));

        memset(degree,0,sizeof(degree));

        for(int i=0;i<=6;i++)

            father[i]= i;

        for(int i=1;i<=n;i++){

            scanf("%d%d",&a,&b);

            vis[a]=1;vis[b]=1;

            add(a,b,i);

            degree[a]++; degree[b]++;

            link(a,b);

        }

        if(judge()==0) printf("No solution\n");

        else{

            dfs(bx);

            while(!q.empty()){

                Node t= q.top(); q.pop();

                if(t.t==1) printf("%d +\n",t.x);

                else printf("%d -\n",t.x);

            }

            printf("\n");

        }

    return 0;

}