Euler Circuit UVA

就是求混合图是否存在欧拉回路

如果存在则输出一组路径

（我就说嘛 咱的代码怎么可能错。。。。。最后的输出格式竟然w了一天 我都没发现）

解析：

　　对于无向边定向建边放到网络流图中add(u, v, 1);

　　对于有向边放到另一个图中add2(u, v);

　　然后就是混合边求是否有欧拉  

　　一边dinic后 遍历每一条边 如果不是反向边 且 起点不是s 终点不是t 

　　如果Node[i].c == 0 则 add2(Node[i].v, Node[i].u);

　　else add2(Node[i].u, Node[i].v);

　　然后用有向图的fleury输出边就好了

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d
", a);
#define plld(a) printf("%lld
", a);
#define pc(a) printf("%c
", a);
#define ps(a) printf("%s
", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 100010, INF = 2e9;
int n, m, s, t, cnt;
int in[maxn], out[maxn], vis[maxn];
int d[maxn], head[maxn], cur[maxn];
set<int> ss;
int st[maxn],cnt3;
int cnt2, head2[maxn];


struct edge
{
    int u, v, c, next, ff;
}Edge[maxn << 1];

void add_(int u, int v, int c, int ff)
{
    Edge[cnt].u = u;
    Edge[cnt].v = v;
    Edge[cnt].c = c;
    Edge[cnt].ff = ff;
    Edge[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c, 1);
    add_(v, u, 0, 0);
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -1; i = Edge[i].next)
        {
            edge e = Edge[i];
            if(!d[e.v] && e.c > 0)
            {
                d[e.v] = d[e.u] + 1;
                Q.push(e.v);
                if(e.v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0;
    if(u == t || cap == 0)
        return cap;
    for(int &i = cur[u]; i != -1; i = Edge[i].next)
    {
        edge e = Edge[i];
        if(d[e.v] == d[u] + 1 && e.c > 0)
        {
            int V = dfs(e.v, min(cap, e.c));
            Edge[i].c -= V;
            Edge[i^1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

int Dinic(int u)
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof(head));
        ans += dfs(u, INF);
    }
    return ans;
}


struct node
{
    int u, v, flag, next;
}Node[maxn << 1];

void add2(int u, int v)
{
    Node[cnt2].u = u;
    Node[cnt2].v = v;
    Node[cnt2].next = head2[u];
    Node[cnt2].flag = 0;
    head2[u] = cnt2++;
}
int used[maxn];
void fleury(int u)
{
    for(int i = head2[u]; i != -1; i = Node[i].next)
    {
        if(!used[i])
        {
            used[i] = 1;
            fleury(Node[i].v);
        }

    }
    st[cnt3++] = u;
}


void init()
{
    mem(in, 0);
    mem(head, -1);
    mem(out, 0);
    mem(st, 0);
    cnt = 0;
    cnt2 = 0;
    cnt3 = 0;
    mem(head2, -1);
    mem(used, 0);
    ss.clear();
    
}

char str[2];

int main()
{
    int T;
    cin >> T;
    while(T--)
    {
        int u, v, w;
        cin >> n >> m;
        init();
        s = 0, t = maxn - 1;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%s", &u, &v, str);
            in[v]++, out[u]++;
            if(str[0] == 'U') add(u, v, 1);
            else if(str[0] == 'D') add2(u, v);
        }
        int flag = 0, m_sum = 0;
        for(int i = 1; i <= n; i++)
        {
            if(abs(out[i] - in[i]) & 1)
            {
                flag = 1;
                break;
            }
            if(out[i] > in[i]) add(s, i, (out[i] - in[i]) / 2), m_sum += (out[i] - in[i]) / 2;
            else if(in[i] > out[i]) add(i, t, (in[i] - out[i]) / 2);

        }
        if(!flag && m_sum == Dinic(s))
        {
            for(int i = 0; i < cnt; i++)
            {
                if(!Edge[i].ff || Edge[i].u == s || Edge[i].v == t) continue;
                if(Edge[i].c == 0) add2(Edge[i].v, Edge[i].u);
                else add2(Edge[i].u, Edge[i].v);
            }
            fleury(1);
            for(int i = cnt3 - 1; i >= 0; i--)
            {
                if(i != cnt3 - 1) printf(" ");
                printf("%d", st[i]);
            }


            printf("
");


        }
        else
            cout << "No euler circuit exist" << endl;
        if(T) printf("
");

    }

    return 0;
}