Poj 1041 John's trip

Euler Loop

A directed graph is used to determine whether it is an Euler loop and the output path. The minimum Lexicographic Order is required. In the output, x y z, x, and y are vertex numbers (up to 44 points) and z are edge numbers (numbered here, the maximum number of edges). The output path is not the output vertex but the number of the output edge. Therefore, the minimum Lexicographic Order is the minimum Lexicographic Order of the edge. Each group of data ends at 0. The first row of each array, with two dots x and y, selects the smaller one as the starting point.

In this case, the graph is connected, so you do not need to judge the connectivity. Therefore, to judge whether it is a euclidean, you only need to check whether the degree of each vertex is an even number. If not, there is no Euler loop, if yes, there is an Euler loop.

The path with the smallest Lexicographic Order is output. You can use a graph creation method.

E [k] [0], e [k] [1] indicates the two vertices of the k edge.

We use recursion to output the path, starting from the start point, then enumeration edge, starting from the smallest side of the number, so as to ensure the minimum Lexicographic Order

 

#include <iostream>#include <cstdio>#include <cstring>#include <stack>using namespace std;#define N 50#define M 2010int node,edge,sp,Max;int e[M][2],de[M];bool used[M];stack<int>sta;void dfs(int u){    for(int k=1; k<=edge; k++)        if(!used[k])        {            int v;            if(e[k][0] == u) v = e[k][1];            else if(e[k][1] == u) v = e[k][0];            else continue;            used[k] = true;            dfs(v);            sta.push(k);        }}void solve(){    while(!sta.empty()) sta.pop();    memset(used,false,sizeof(used));    dfs(sp);    bool first = true;    while(!sta.empty())    {        if(!first) cout << " ";        cout << sta.top();        first = false;        sta.pop();    }    cout << endl;}int main(){    int u,v,m;    while(cin >> u >> v)    {        if(u == 0 && v == 0) break;        memset(de,0,sizeof(de));        sp = min(u,v); Max = max(u,v);        cin >> m;        e[m][0] = u;  e[m][1] = v;        de[u]++;  de[v]++;        edge = 1;        while(cin >> u >> v)        {            if(u == 0 && v == 0) break;            Max = max(Max,u); Max = max(Max,v);            cin >> m;            e[m][0] = u; e[m][1] = v;            de[u]++; de[v]++;            edge++;        }        int ok = 1;        for(int i=1; i<=Max; i++)             if(de[i] & 1)            { ok = 0; break;}        if(!ok)        {            cout << "Round trip does not exist." << endl;            continue;        }        solve();    }    return 0;}