[SCU - 4520 Euler]欧拉通路的判定

题目链接：[SCU - 4520 Euler] 

解题思路：

1. 无向图存在欧拉回路的充要条件：连通且没有奇度顶点。
2. 无向图存在欧拉路径的充要条件：连通且奇度顶点个数为2。
3. 有向图存在欧拉路径的充要条件：基图连通且存在某顶点入度比出度大1，另一顶点出度比入度大1，其余顶点入度等于出度。
4. 有向图存在欧拉回路的充要条件：基图连通且所有顶点入度等于出度。

有向图的连通性可以用并查集来判断，而无向图的连通性可以拓扑排序来判断。但是， 这个题目竟然有向图也能用并查集来判断，也是有毒了... 

#include <bits/stdc++.h>using namespace std;const int MAXN = 1000 + 5;int T, N, M;int deg[MAXN];int par[MAXN];int in[MAXN], out[MAXN];int Find (int x) {    return par[x] == -1 ? x : par[x] = Find (par[x]);}void Union (int a, int b) {    int pa = Find (a), pb = Find (b);    if (pa != pb) par[pb] = pa;}void init() {    memset (deg, 0, sizeof (deg) );    memset (in, 0, sizeof (in) );    memset (out, 0, sizeof (out) );    memset (par, -1, sizeof (par) );}bool connected() {    int cnt = 0;    for (int i = 1; i <= N; i++) {        if (par[i] == -1) cnt ++;    }    return cnt <= 1;}bool dir() {    int num;    num = 0;    for (int i = 1; i <= N; i++) {        if (deg[i] & 1) num ++;    }    return (num == 0 || num == 2);}bool undir() {    int num, s, t;    num = 0;    s = t = -1;    for (int i = 1; i <= N; i++) {        if(out[i] > in[i]) s = i, num ++;        else if(in[i] > out[i]) t = i, num ++;    }    if(!(num == 0 || (num == 2 && ~s && ~t && (out[s] - in[s]) == 1) && (out[t] - in[t]== -1))) return false;}int main() {///    freopen ("input.txt", "r", stdin);    int u, v;    scanf("%d", &T);    while (T --) {        init();        scanf ("%d %d", &N, &M);        for (int i = 1; i <= M; i++) {            scanf ("%d %d", &u, &v);            deg[u] ++, deg[v] ++;            Union (u, v);            out[u] ++, in[v] ++;        }        bool con, res1, res2;        con = connected();        if(!con) {            puts("No No");            continue;        }        res1 = dir();        res2 = undir();        printf("%s %s\n", res1 ? "Yes" : "No", res2 ? "Yes" : "No");    }    return 0;}