POJ 1041 求解欧拉回路方案

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这题模板太神了……几乎不用自己再写什么，只要把图建好就行了……

对了，刚才做了后琦神说了欧拉回路了汉密顿回路才记得其区别：欧拉回路是一笔画问题，即边走且只走一次；而汉密顿回路是点走且只下次一次。

#include <iostream>#include <cstdio>#include <fstream>#include <algorithm>#include <cmath>#include <deque>#include <vector>#include <list>#include <queue>#include <string>#include <cstring>#include <map>#include <stack>#include <set>#define PI acos(-1.0)#define mem(a,b) memset(a,b,sizeof(a))#define sca(a) scanf("%d",&a)#define sc(a,b) scanf("%d%d",&a,&b)#define pri(a) printf("%d\n",a)#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define MM 1000005#define MN 2000#define INF 1000000009#define eps 1e-7using namespace std;typedef long long ll;int f[MN];vector<int>path;vector< pair<int,int> >adj[MN];bool vis[MM];int find(int x){    return x==f[x]?x:f[x]=find(f[x]);}void add(int x,int y,int z){    adj[x].push_back(make_pair(z,y));    adj[y].push_back(make_pair(z,x));}void dfs(int u){    for(int i=0; i<adj[u].size(); i++)        if(!vis[adj[u][i].first])        {            vis[adj[u][i].first]=true;            dfs(adj[u][i].second);            path.push_back(adj[u][i].first);        }}bool eluer(){    int i,j,origin=-1;    for(i=0; i<MN; i++) f[i]=i;    for(i=0; i<MN; i++)        for(j=0; j<adj[i].size(); j++)            f[find(i)]=find(adj[i][j].second);    for(i=0; i<MN; i++)        if(adj[i].size())        {            if(adj[i].size()%2==1) return false;            if(origin==-1) origin=i;            if(find(i)!=find(origin)) return false;            sort(adj[i].begin(),adj[i].end());        }    path.clear();    mem(vis,false);    if(origin!=-1) dfs(origin);    reverse(path.begin(),path.end());    return true;}int main(){    int x,y,z,i,len;    while(sc(x,y)&&(x||y))    {        for(i=0;i<MN;i++) adj[i].clear();        sca(z),add(x,y,z);        while(sc(x,y)&&(x||y))            sca(z),add(x,y,z);        if(eluer())        {            len=path.size();            for(i=0;i<len;i++)                printf("%d%c",path[i],i!=len-1?' ':'\n');        }        else puts("Round trip does not exist.");    }    return 0;}

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