SGU101 - Domino(欧拉路)

思路：

把牌的两个数值抽象成有两个端点的路径，然后找到一个欧拉路径能够遍历所有的边即可。

代码如下：

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
#define M 105
vector<int>g[10][10];
bool vis[M];
int n, du[10], ans[M];
int dfs(int cur, int x)
{
if(cur==n+1) return 1;
for(int i = 0; i <= 6; ++i)
{
for(int j = 0; j < (int)g[x][i].size(); ++j)
{
int l = g[x][i][j];
l = l<0?-1*l:l;
if(vis[l]==0)
{
vis[l] = 1;
ans[cur] = g[x][i][j];
if(dfs(cur+1,i)) return 1;
vis[l] = 0;
break;
}
}
}
return 0;
}
int main ()
{
int a, b;
scanf("%d",&n);
for(int i = 1; i <= n; ++i)
{
scanf("%d %d",&a,&b);
g[a][b].push_back(i);
g[b][a].push_back(-i);
++du[a];
++du[b];
}
int st = 0, cnt = 0;
for(int i = 0; i <= 6; ++i)
if(du[i]) { st = i; break; }
for(int i = 0; i <= 6; ++i)
{ if(du[i]&1) {st = i; ++cnt;} }
if(cnt!=0&&cnt!=2) { printf("No solution\n"); return 0; }
int f = dfs(1,st);
if(f)
{
for(int i = 1; i <= n; ++i)
printf("%d %c\n",(ans[i]<0?-1*ans[i]:ans[i]), (ans[i]<0?'-':'+'));
}
else printf("No solution\n");
return 0;
}