HS BDC HDU - 3472（混合欧拉路径）

题意：

　　就是混合欧拉路径板题

解析：

　　欧拉路径加一条t_ ---> s_  的边就变成了欧拉回路，所以利用这一点，如果存在两个奇点，那么这两个奇点出度大的是s_，入度大的是t_，加一条t_ ---> s_的容量为1的边即可

然后混合欧拉回路板题

#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\n", a);
#define plld(a) printf("%lld\n", a);
#define pc(a) printf("%c\n", a);
#define ps(a) printf("%s\n", 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 = , INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m, s, t, cnt;
int f[maxn], deg[maxn], in[maxn], out[maxn], vis[maxn];
int d[maxn], head[maxn], cur[maxn];
set<int> ss;

int find(int x)
{
    return f[x] == x ? x : (f[x] = find(f[x]));
}

void init()
{
    for(int i = ; i < maxn; i++) f[i] = i;
    mem(in, );
    mem(head, -);
    mem(out, );
    cnt = ;
    mem(vis, );
    ss.clear();
}

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

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

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

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

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

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

int main()
{
    int T, kase = ;
    cin >> T;
    while(T--)
    {
        string str;
        int u, v, w;
        cin >> n;
        init();
        s = , t = ;
        for(int i = ; i <= n; i++)
        {
            cin >> str >> w;
            u = str[] - 'a' + , v = str[str.size() - ] - 'a' + ;
            vis[u] = vis[v] = ;
            in[v]++, out[u]++;
            if(u != v && w == ) add(u, v, );
            int l = find(u);
            int r = find(v);
            if(l != r) f[l] = r;
        }
        int flag = , m_sum = , s_ = INF, t_ = INF;
        for(int i = ; i <= ; i++)
        {
            int x = find(i);
            if(vis[x]) ss.insert(x);
            if(abs(out[i] - in[i]) & )
            {
                if(out[i] > in[i]) s_ = i;
                else t_ = i;
                if(++flag > ) break;
            }
            if(out[i] > in[i]) add(s, i, (out[i] - in[i]) / ), m_sum += (out[i] - in[i]) / ;
            else if(in[i] > out[i]) add(i, t, (in[i] - out[i]) / );

        }
        if(s_ != INF && t_ != INF) add(t_, s_,  );
     //   cout << flag << "   " << ss.size() << endl;
        printf("Case %d: ", ++kase);
        if(flag >  || ss.size() >  || flag == )
        {
            cout << "Poor boy!" << endl;
            continue;
        }
        if(m_sum == Dinic(s))
            cout << "Well done!" << endl;
        else
            cout << "Poor boy!" << endl;

    }

    return ;
}