csu1617]强连通分量

题意：定义域属于一个集合S={0,1,...,n-1}，求S的子集个数，满足以子集的元素为定义域的函数P(x)的值域等于子集本身。

思路：以元素为点，x到P(x)连一条有向边，不难发现，如果有一个有向环，那么环上的元素构成的集合就满足要求。所以问题转化为求有向环的个数，由于有向环之间不可能有交点（同一个点有且仅有一条出边），所以答案就是2^有向环的个数（如果选了有向环上的一点，那么整个有向环必须全部选）。所以只要用tarjan算法统计点数大于等于2的强连通分量个数然后加上自环的，就得到了有向环的个数了。

  1 #pragma comment(linker, "/STACK:10240000,10240000")

  2  

  3 #include <iostream>

  4 #include <cstdio>

  5 #include <algorithm>

  6 #include <cstdlib>

  7 #include <cstring>

  8 #include <map>

  9 #include <queue>

 10 #include <deque>

 11 #include <cmath>

 12 #include <vector>

 13 #include <ctime>

 14 #include <cctype>

 15 #include <stack>

 16 #include <set>

 17 #include <bitset>

 18 #include <functional>

 19 #include <numeric>

 20 #include <stdexcept>

 21 #include <utility>

 22  

 23 using namespace std;

 24  

 25 #define mem0(a) memset(a, 0, sizeof(a))

 26 #define mem_1(a) memset(a, -1, sizeof(a))

 27 #define lson l, m, rt << 1

 28 #define rson m + 1, r, rt << 1 | 1

 29 #define define_m int m = (l + r) >> 1

 30 #define rep_up0(a, b) for (int a = 0; a < (b); a++)

 31 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)

 32 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)

 33 #define rep_down1(a, b) for (int a = b; a > 0; a--)

 34 #define all(a) (a).begin(), (a).end()

 35 #define lowbit(x) ((x) & (-(x)))

 36 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}

 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}

 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

 39 #define pchr(a) putchar(a)

 40 #define pstr(a) printf("%s", a)

 41 #define sstr(a) scanf("%s", a)

 42 #define sint(a) scanf("%d", &a)

 43 #define sint2(a, b) scanf("%d%d", &a, &b)

 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)

 45 #define pint(a) printf("%d\n", a)

 46 #define test_print1(a) cout << "var1 = " << a << endl

 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl

 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

 49 #define mp(a, b) make_pair(a, b)

 50 #define pb(a) push_back(a)

 51  

 52 typedef long long LL;

 53 typedef pair<int, int> pii;

 54 typedef vector<int> vi;

 55  

 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};

 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };

 58 const int maxn = 1e4 + 7;

 59 const int md = 1e9 + 7;

 60 const int inf = 1e9 + 7;

 61 const LL inf_L = 1e18 + 7;

 62 const double pi = acos(-1.0);

 63 const double eps = 1e-6;

 64  

 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}

 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}

 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}

 68 template<class T>T condition(bool f, T a, T b){return f?a:b;}

 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}

 70 int make_id(int x, int y, int n) { return x * n + y; }

 71  

 72 struct Graph {

 73     vector<vector<int> > G;

 74     void clear() { G.clear(); }

 75     void resize(int n) { G.resize(n + 2); }

 76     void add(int u, int v) { G[u].push_back(v); }

 77     vector<int> & operator [] (int u) { return G[u]; }

 78 };

 79 Graph G;

 80 int n, m;

 81 int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt;

 82 stack<int> S;

 83 int cnt[maxn], a[maxn];

 84  

 85 void dfs(int u) {

 86     pre[u] = lowlink[u] = ++ dfs_clock;

 87     S.push(u);

 88     rep_up0(i, G[u].size()) {

 89         int v = G[u][i];

 90         if (!pre[v]) {

 91             dfs(v);

 92             min_update(lowlink[u], lowlink[v]);

 93         }

 94         else if (!sccno[v]) {

 95             min_update(lowlink[u], pre[v]);

 96         }

 97     }

 98     if (lowlink[u] == pre[u]) {

 99         scc_cnt ++;

100         for(;; ) {

101             int x = S.top(); S.pop();

102             sccno[x] = scc_cnt;

103             if (x == u) break;

104         }

105     }

106 }

107 int find_scc(int n) {

108     dfs_clock = scc_cnt = 0;

109     mem0(sccno);

110     mem0(pre);

111     rep_up0(i, n) {

112         if (!pre[i]) dfs(i);

113     }

114     mem0(cnt);

115     int c = 0;

116     rep_up0(i, n) {

117         cnt[sccno[i]] ++;

118     }

119     rep_up1(i, scc_cnt) {

120         if (cnt[i] >= 2) c ++;

121     }

122     rep_up0(i, n) if (G[i].size() == 0) c ++;

123     int ans = 1;

124     rep_up0(i, c) {

125         ans = (ans << 1) % md;

126     }

127     return ans;

128 }

129  

130 int P(int x) {

131     int ans = 0;

132     rep_up0(i, m + 1) {

133         ans = (ans * x + a[m - i]) % n;

134     }

135     return ans;

136 }

137  

138 int main() {

139     //freopen("in.txt", "r", stdin);

140     int T;

141     cin >> T;

142     while (T --) {

143         cin >> n >> m;

144         G.clear();

145         G.resize(n);

146         rep_up0(i, m + 1) sint(a[i]);

147         rep_up0(i, n) {

148             int x = P(i);

149             if (x != i) G.add(i, x);

150         }

151         cout << find_scc(n) << endl;

152     }

153     return 0;

154 }

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