HDU-1878 欧拉回路 欧拉回路

题目链接：https://cn.vjudge.net/problem/HDU-1878

题意

中文题，而且就是单纯的欧拉回路

思路

1. 判断连通图
   用并查集会很好，bfs亦可
   一时脑抽用bfs过了这个题，数据还是太弱
2. 出度==入度

代码

并查集查连通

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=1000;
struct Node{
int parent, rank;
Node(int parent=0, int rank=0):
parent(parent), rank(rank) {}
}node[maxn+5];
int n;
int find(int x){
return (node[x].parent==x)?x:(node[x].parent=find(node[x].parent));
}

void join(int a, int b){
a=find(a); b=find(b);
if (a==b) return;
if (node[a].rank==node[b].rank) node[a].rank++;
if (node[a].rank>node[b].rank) node[b].parent=a;
else node[a].parent=b;
}

bool connect(void){
for (int i=2; i<=n; i++)
if (find(1)!=find(i)) return false;
return true;
}

int main(void){
while (scanf("%d", &n)==1 && n){
int m, cnt=0, vis[maxn+5]={0};
bool set[maxn+5]={false};
for (int i=1; i<=n; i++) node[i]=Node(i, 0);

scanf("%d", &m);
for (int i=0, a, b; i<m; i++){
scanf("%d%d", &a, &b);
join(a, b);
vis[a]++; vis[b]++;
}

int flag=false;
for (int i=1; i<=n; i++)
if (vis[i]%2) {flag=true; break;}
if (flag==false && !connect()) flag=true;
printf("%d\n", (flag)?0:1);
}

return 0;
}

BFS查连通

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=1000;
struct Edge{
int from, to;
bool vis;
Edge(int from=0, int to=0, int vis=false):
from(from), to(to), vis(vis) {}
};
vector<Edge> edge;
vector<int> G[maxn+5];
int n;
inline void addEdge(int from, int to){
edge.push_back(Edge(from, to, false));
G[from].push_back(edge.size()-1);
G[to].push_back(edge.size()-1);
}

bool connective(void){
int cnt=1; bool vis[maxn+5]={false};
queue<int> que;
que.push(1); vis[1]=true;
while (que.size()){
int from=que.front(); que.pop();
if (cnt==n) return true;
for (int i=0; i<G[from].size(); i++){
Edge &e=edge[G[from][i]];
int to=(e.to==from)?e.from:e.to;
if (e.vis) continue;
vis[to]=true; e.vis=true; cnt++;
que.push(to);
}
}return false;
}

int main(void){
while (scanf("%d", &n)==1 && n){
int m, vis[maxn+5]={0};
memset(G, 0, sizeof(G));

scanf("%d", &m);
for (int i=0, a, b; i<m; i++){
scanf("%d%d", &a, &b);
addEdge(a, b);// G[a][b]++; G[b][a]++;
vis[a]++; vis[b]++;
}

int flag=false;
for (int i=1; i<=n; i++)
if (vis[i]%2) {flag=true; break;}
if (flag==false && connective()==0) flag=true;
printf("%d\n", (flag)?0:1);
}

return 0;
}

并查集

BFS