SGU 122
题意：给你n个人，每个人有大于 N / 2(向上取整)的朋友，问你1这个人有一个书，每个人都想看，只能从朋友之间传递，然后最后回到了1这个人，问你
是否有解，然后有解输出路径
收获：哈密尔顿路
一:Dirac定理(充分条件)
设一个无向图中有N个顶点,若所有顶点的度数大于等于N/2,则哈密顿回路一定存在.(N/2指的是⌈N/2⌉,向上取整)
二:基本的必要条件
设图G=<V, E>是哈密顿图,则对于v的任意一个非空子集S,若以|S|表示S中元素的数目,G-S表示G中删除了S中的点以及这些点所关联的边后得到的子图,则W(G-S)<=|S|成立.其中W(G-S)是G-S中联通分支数.
三:竞赛图(哈密顿通路)
N(N>=2)阶竞赛图一点存在哈密顿通路.
还偷了一个哈密尔顿回路模板：https://blog.csdn.net/u010929036/article/details/46345059
而且这道题会卡输入的，我用getline超时了。。。
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e3+6; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } class Hamilton { int n, next[maxn]; bool g[maxn][maxn], vis[maxn]; int find(int u) { for (int v = 0; v < n; ++v) if (g[u][v] && !vis[v]) return v; return -1; } void reverse(int v, int f) { if (v == -1) return; reverse(next[v], v); next[v] = f; } public: void init(int n) { this->n = n; memset(g, false, sizeof(g)); } void add_edge(int u, int v) { g[u][v] = true; } void find_path(int s = 0) { int t = s, sz = 1; memset(next, -1, sizeof(next)); memset(vis, false, sizeof(vis)); vis[s] = true; while (sz < n) { if (sz == 1) { for (int v; ~(v = find(s)); s = v) ++sz, vis[v] = true, next[v] = s; for (int v; ~(v = find(t)); t = v) ++sz, vis[v] = true, next[t] = v; } else { for (int u, v = 0; v < n; ++v) if (!vis[v]) { ++sz, vis[v] = true; for (u = s; !g[u][v]; u = next[u]); s = next[u]; t = next[u] = v; break; } } if (g[t][s]) next[t] = s; for (int u = next[s], v; next[t] == -1; u = next[u]) if (g[u][t] && g[v=next[u]][s]) reverse(v, s), next[u] = t, t = v; } for (int i = 0, u = 0; i < n; ++i, u = next[u]) printf("%d ", u + 1); printf("%d\n", 1); } } grp; char s[100011],*p; //s可以对应char *，而不能对应char* & bool get_int(int &v,char* &p){ v = 0; while(*p && !isdigit(*p)) p++; if(!isdigit(*p)) return false; while(isdigit(*p)) v = v * 10 + *p++ - '0'; return true; } int main(){ int n,v; scanf("%d",&n);getchar(); grp.init(n); rep(i,0,n){ gets(s);p = s; // de(s) while(get_int(v,p)) grp.add_edge(i,--v); } grp.find_path(); return 0; }