BZOJ4337 树的同构 (树哈希)（未完成）

样例迷，没过

交了30pts
#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

#include <cmath>

#define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a))

#define nR(a,b,c) for(register int a = (b); (a) >= (c); --(a))

#define Fill(a,b) memset(a, b, sizeof(a))

#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))

#define ll long long

#define u32 unsigned int

#define u64 unsigned long long

#define ON_DEBUGG

#ifdef ON_DEBUGG

#define D_e_Line printf("\n----------\n")

#define D_e(x) cout << (#x) << " : " << x << endl

#define Pause() system("pause")

#define FileOpen() freopen("in.txt", "r", stdin)

#define FileSave() freopen("out.txt", "w", stdout)

#include <ctime>

#define TIME() fprintf(stderr, "\ntime: %.3fms\n", clock() * 1000.0 / CLOCKS_PER_SEC)

#else

#define D_e_Line ;

#define D_e(x) ;

#define Pause() ;

#define FileOpen() ;

#define FileSave() ;

#define TIME() ;

//char buf[1 << 21], *p1 = buf, *p2 = buf;

//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)

#endif

using namespace std;

struct ios{

    template<typename ATP>inline ios& operator >> (ATP &x){

        x = 0; int f = 1; char ch;

        for(ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;

        while(ch >= '0' && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();

        x *= f;

        return *this;

    }

}io;

template<typename ATP>inline ATP Max(ATP a, ATP b){

    return a > b ? a : b;

}

template<typename ATP>inline ATP Min(ATP a, ATP b){

    return a < b ? a : b;

}

template<typename ATP>inline ATP Abs(ATP a){

    return a < 0 ? -a : a;

}

#include <climits>

const int N = 1003;

const int MOD = 9999991;

#define int long long

struct Edge{

	int nxt, pre;

}e[N];

int head[N], cntEdge;

inline void add(int u, int v) {

	e[++cntEdge] = (Edge){ head[u], v}, head[u] = cntEdge;

}

int prime[1013], primeIndex;

bool vis[8007];

inline void EularPhi(int n) {

	R(i,2,n){

		if(!vis[i]) prime[++primeIndex] = i;

		for(register int j = 1; j <= primeIndex && i * prime[j] <= n; ++j){

			vis[i * prime[j]] = 1;

			if(i % prime[j] == 0) break;

		}

	}

}

int siz[N];

int f[N], g[N], fa[N];

inline void DFS(int u, int father){

	siz[u] = 1;

	f[u] = 1;

	fa[u] = father;

	for(register int i = head[u]; i; i = e[i].nxt){

		int v = e[i].pre;

		if(v == father) continue;

		DFS(v, u);

		siz[u] += siz[v];

		f[u] = (f[u] + f[v] * prime[siz[v]] % MOD + MOD) % MOD;

	}

}

namespace HASH{

	struct Node{

		int nxt, pre, w;

	}e[N * N];

	int head[MOD + 3], cntHash;

	struct Hash{

		inline void Insert(int x, int id){

			int u = x % MOD;

			for(register int i = head[u]; i; i = e[i].nxt){

				int v = e[i].pre;

				if(v == x){

					return;

				}

			}

			e[++cntHash] = (Node){ head[u], x, id}, head[u] = cntHash;

		}

		inline int Query(int x){

			int u = x % MOD;

			for(register int i = head[u]; i;  i= e[i].nxt){

				int v = e[i].pre;

				if(v == x){

					return e[i].w;

				}

			}

			return 0;

		}

	}H;

}

#undef int

int main(){

#define int long long

	EularPhi(8000);

	int n, m;

	io >> m;

	R(id,1,m){

		cntEdge = 0;

		Fill(head, 0);

		io >> n;

		R(i,1,n){

			int fa;

			io >> fa;

			add(i, fa);

			add(fa, i);

		}

		DFS(0, 0);

		int sum = LLONG_MAX;

		R(i,0,n){

			DFS(i, i);

			sum = Min(sum, f[i]);

		}

		HASH::H.Insert(sum, id);

		printf("%lld\n", HASH::H.Query(sum));

	}

	return 0;

}
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BZOJ4337 树的同构 (树哈希)（未完成）

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