[SDOI2007]游戏【哈希+DAG拓扑】

题目链接

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先通过哈希确定点，这里我使用的是双值哈希。然后利用哈希判断可以和前面的出现的点如何链接。

之后构造出来的图一定是一副DAG图，有向无环图，所以直接拓扑排序DP即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const ull h1 = 29, h2 = 31;
const int maxN = 1e4 + 7;
int N = 0, num[26];
ull b1[27], b2[27];
char s[maxN][105];
map<pair<ull, ull>, int> mp;
pair<ull, ull> id[maxN];
void Pre_did()
{
    b1[0] = b2[0] = 1;
    for(int i=1; i<26; i++)
    {
        b1[i] = b1[i - 1] * h1;
        b2[i] = b2[i - 1] * h2;
    }
}
int head[maxN], cnt = 0, du[maxN];
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN << 1];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
int dp[maxN], pre[maxN], ans_val, ans_id;
queue<int> Q;
inline void tp()
{
    while(!Q.empty()) Q.pop();
    for(int i=1; i<=N; i++) if(!du[i]) { Q.push(i); dp[i] = 1; pre[i] = 0; ans_val = 1; ans_id = i; }
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to;
            if(dp[u] + 1 > dp[v])
            {
                dp[v] = dp[u] + 1;
                pre[v] = u;
            }
            if(dp[v] > ans_val)
            {
                ans_val = dp[v];
                ans_id = v;
            }
            du[v]--;
            if(!du[v]) Q.push(v);
        }
    }
}
int main()
{
    Pre_did();
    while(scanf("%s", s[++N]) != EOF)
    {
        du[N] = 0; head[N] = -1;
        int len = (int)strlen(s[N]);
        memset(num, 0, sizeof(num));
        for(int i=0; i<len; i++) num[s[N][i] - 'a']++;
        ull hash_val_1 = 0, hash_val_2 = 0, tmp_1, tmp_2;
        for(int i=0; i<26; i++)
        {
            hash_val_1 = hash_val_1 + (ull)num[i] * b1[i];
            hash_val_2 = hash_val_2 + (ull)num[i] * b2[i];
        }
        id[N] = MP(hash_val_1, hash_val_2);
        mp[id[N]] = N;
        for(int i=0; i<26; i++)
        {
            tmp_1 = hash_val_1 - b1[i]; tmp_2 = hash_val_2 - b2[i];
            if(mp[MP(tmp_1, tmp_2)])
            {
                du[mp[MP(tmp_1, tmp_2)]]++;
                addEddge(N, mp[MP(tmp_1, tmp_2)]);
            }
            tmp_1 = hash_val_1 + b1[i]; tmp_2 = hash_val_2 + b2[i];
            if(mp[MP(tmp_1, tmp_2)])
            {
                du[N]++;
                addEddge(mp[MP(tmp_1, tmp_2)], N);
            }
        }
        if(!mp[MP(hash_val_1, hash_val_2)]) mp[MP(hash_val_1, hash_val_2)] = N;
    }
    N--;
    tp();
    printf("%d\n", ans_val);
    while(ans_id)
    {
        printf("%s\n", s[ans_id]);
        ans_id = pre[ans_id];
    }
    return 0;
}