Uva10054_the neck.pdf

It's easy to find the Euler loop. And output.

Only focus on using stacks to control output.

Why,

If the stack is not used, 1-> 2-> 3-> 1 is returned, and 4-> 5 is output. The connection fails, so if the stack is used, A complete path will be saved.

Because it is an undirected graph, as long as the degree of each point is even, there must be a legal Euler loop.

 

Summon code:

 

 

#include <iostream>#include <cstring>#include <cstdio>using namespace std;int a[51][51],n=50,d[51],m,T;void dfs(int x){    for (int i=1; i<=n; i++)        if (a[x][i])        {            a[x][i]--,a[i][x]--;            dfs(i);            printf("%d %d\n",i,x);        }}int main(){    int cas=0,U,V;    scanf("%d",&T);    while (T--)    {        memset(d,0,sizeof d);        memset(a,0,sizeof a);        scanf("%d",&m);        while (m--)        {            scanf("%d%d",&U,&V);            a[U][V]++,a[V][U]++;            d[U]++,d[V]++;        }        bool ans=true;        for (int i=1; i<=n; i++)            if (d[i]&1) ans=false;        if (cas) printf("\n");        printf("Case #%d\n",++cas);        if (!ans) printf("some beads may be lost\n");            else dfs(U);    }    return 0;}