bzoj4205 卡牌配对

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卡牌配对

题目背景：

bzoj4205

分析：我相信很多人第一反应都是暴力连边然后暴力二分图匹配，然后我的常数还没有小到n²过30000，所以我们来想想该怎么办，考虑到200以内的质数只有46个，我们可以选择增加3 * 46 * 46个点，将左边的某种属性有第i个质因数，另一个属性有第j个质因数的点连向对应的一个点，并将这个点与右边满足上述条件的点相连，然后新建源点汇点就可以跑最大流了，大概是一个70000个点，2000000条边的网络流（200以内的数最多3个质因数，所以最多不超过27条边），他竟然让大胆玩高渐进复杂度，那当然就直接上吧。

Source：

/*    created by scarlyw*/#include <cstdio>#include <string>#include <algorithm>#include <cstring>#include <iostream>#include <cmath>#include <cctype>#include <vector>#include <set>#include <queue>  inline char read() {    static const int IN_LEN = 1024 * 1024;    static char buf[IN_LEN], *s, *t;    if (s == t) {        t = (s = buf) + fread(buf, 1, IN_LEN, stdin);        if (s == t) return -1;    }    return *s++;}  ///*template<class T>inline void R(T &x) {    static char c;    static bool iosig;    for (c = read(), iosig = false; !isdigit(c); c = read()) {        if (c == -1) return ;        if (c == '-') iosig = true;     }    for (x = 0; isdigit(c); c = read())         x = ((x << 2) + x << 1) + (c ^ '0');    if (iosig) x = -x;}//*/  const int OUT_LEN = 1024 * 1024;char obuf[OUT_LEN], *oh = obuf;inline void write_char(char c) {    if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;    *oh++ = c;}  template<class T>inline void W(T x) {    static int buf[30], cnt;    if (x == 0) write_char('0');    else {        if (x < 0) write_char('-'), x = -x;        for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;        while (cnt) write_char(buf[cnt--]);    }}  inline void flush() {    fwrite(obuf, 1, oh - obuf, stdout);}  /*template<class T>inline void R(T &x) {    static char c;    static bool iosig;    for (c = getchar(), iosig = false; !isdigit(c); c = getchar())        if (c == '-') iosig = true;     for (x = 0; isdigit(c); c = getchar())         x = ((x << 2) + x << 1) + (c ^ '0');    if (iosig) x = -x;}//*/ const int MAXN = 70000 + 10;const int MAXX = 200 + 5;const int prime[50] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,                        47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103,                        107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163,                        167, 173, 179, 181, 191, 193, 197, 199}; struct node {    int to, w, rev;    node(int to = 0, int w = 0, int rev = 0) : to(to), w(w), rev(rev) {}} ; struct data {    int a, b, c;} x[MAXN], y[MAXN]; std::vector<node> edge[MAXN]; int n1, n2, n, s, t;int id[3][MAXX][MAXX], dis[MAXN], gap[MAXN]; inline void add_edge(int x, int y, int w) {    edge[x].push_back(node(y, w, edge[y].size()));    edge[y].push_back(node(x, 0, edge[x].size() - 1));} inline void read_in() {    R(n1), R(n2);    for (int i = 1; i <= n1; ++i) R(x[i].a), R(x[i].b), R(x[i].c);    for (int i = 1; i <= n2; ++i) R(y[i].a), R(y[i].b), R(y[i].c);} inline int solve(int n, int *a) {    int top = 0;    for (int i = 2; i <= sqrt(n); ++i) {        if (n % i == 0) a[++top] = i;        while (n % i == 0) n /= i;          }    if (n != 1) a[++top] = n;    return top;} inline void build_graph() {    n = n1 + n2;    for (int i = 0; i < 3; ++i)        for (int j = 1; j <= 46; ++j)            for (int k = 1; k <= 46; ++k)                id[i][prime[j]][prime[k]] = ++n;         s = ++n, t = ++n;    int a[MAXX], b[MAXX], c[MAXX];    for (int i = 1; i <= n1; ++i) {        add_edge(s, i, 1);        int top1 = solve(x[i].a, a);        int top2 = solve(x[i].b, b);        int top3 = solve(x[i].c, c);        for (int k1 = 1; k1 <= top1; ++k1)            for (int k2 = 1; k2 <= top2; ++k2)                add_edge(i, id[0][a[k1]][b[k2]], 1);        for (int k1 = 1; k1 <= top1; ++k1)            for (int k3 = 1; k3 <= top3; ++k3)                add_edge(i, id[1][a[k1]][c[k3]], 1);        for (int k2 = 1; k2 <= top2; ++k2)            for (int k3 = 1; k3 <= top3; ++k3)                add_edge(i, id[2][b[k2]][c[k3]], 1);    }    for (int i = 1; i <= n2; ++i) {        add_edge(i + n1, t, 1);        int top1 = solve(y[i].a, a);        int top2 = solve(y[i].b, b);        int top3 = solve(y[i].c, c);        for (int k1 = 1; k1 <= top1; ++k1)            for (int k2 = 1; k2 <= top2; ++k2)                add_edge(id[0][a[k1]][b[k2]], n1 + i, 1);        for (int k1 = 1; k1 <= top1; ++k1)            for (int k3 = 1; k3 <= top3; ++k3)                add_edge(id[1][a[k1]][c[k3]], n1 + i, 1);        for (int k2 = 1; k2 <= top2; ++k2)            for (int k3 = 1; k3 <= top3; ++k3)                add_edge(id[2][b[k2]][c[k3]], n1 + i, 1);    }} ///*inline int sap(int cur, int flow, int s, int t, int n) {    if (cur == t) return flow;    int del = 0;    static int temp[MAXN];    for (int p = temp[cur]; p < edge[cur].size(); ++p) {        node *e = &edge[cur][p];        if (e->w > 0 && dis[e->to] + 1 == dis[cur]) {            int ret = sap(e->to, std::min(flow - del, e->w), s, t, n);            e->w -= ret, edge[e->to][e->rev].w += ret, temp[cur] = p;            if ((del += ret) == flow || dis[cur] >= n)                 return temp[cur] = 0, del;        }    }    if (--gap[dis[cur]] == 0) dis[s] = n;    return gap[++dis[cur]]++, temp[cur] = 0, del;} inline int sap(int s, int t, int n) {    int ret = 0;    for (gap[0] = n; dis[s] < n; ) ret += sap(s, ~0u >> 1, s, t, n);    return ret;}//*/int main() {    read_in();    build_graph();    std::cout << sap(s, t, n);    return 0;}

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