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 = 10010, 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 = 0; i < maxn; i++) f[i] = i; mem(in, 0); mem(head, -1); mem(out, 0); cnt = 0; mem(vis, 0); 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, 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; } int main() { int T, kase = 0; cin >> T; while(T--) { string str; int u, v, w; cin >> n; init(); s = 0, t = 30; for(int i = 1; i <= n; i++) { cin >> str >> w; u = str[0] - 'a' + 1, v = str[str.size() - 1] - 'a' + 1; vis[u] = vis[v] = 1; in[v]++, out[u]++; if(u != v && w == 0) add(u, v, 1); int l = find(u); int r = find(v); if(l != r) f[l] = r; } int flag = 0, m_sum = 0, s_ = INF, t_ = INF; for(int i = 1; i <= 26; i++) { int x = find(i); if(vis[x]) ss.insert(x); if(abs(out[i] - in[i]) & 1) { if(out[i] > in[i]) s_ = i; else t_ = i; if(++flag > 2) 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(s_ != INF && t_ != INF) add(t_, s_, 1); // cout << flag << " " << ss.size() << endl; printf("Case %d: ", ++kase); if(flag > 2 || ss.size() > 1 || flag == 1) { cout << "Poor boy!" << endl; continue; } if(m_sum == Dinic(s)) cout << "Well done!" << endl; else cout << "Poor boy!" << endl; } return 0; }