Bzoj4205卡牌配对

原题网址：http://www.lydsy.com/JudgeOnline/problem.php?id=4205
大致模型是二分图最大匹配，但暴力建边会T，考虑建中间节点，至少要有两项属性值不互质，但这样中间点数还是爆炸和边数，然而看了题解才知道中间点只要质数即可。（一开始反向边流量又忘记设为0了。。。
%hzwer http://hzwer.com/7428.html

#include<bits/stdc++.h>const int N = 7e4;const int M = 2e6;const int INF = 1e9;template <typename T> void read(T &x){    x = 0; T f = 1;    char c = getchar();    while (c < '0' || c > '9') {if (c == '-') f *= -1; c = getchar();}    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}    x *= f;}struct edge{int y,v,next;} mp[M*2];int n1,n2,S,T,first[N],cur[N],p[N],a[N],b[N],c[N],h[N],q[N],x[N],y[N],z[N],s,cnt,ans;void ins(int x, int y, int v){    mp[++s] = (edge) {y,v,first[x]}; first[x] = s;    mp[++s] = (edge) {x,0,first[y]}; first[y] = s;}bool isprime(int n){    if (n == 1) return 0;    for (int i=2; i*i<=n; i++)        if (n % i == 0)            return 0;    return 1;}void init(){    for (int i=1; i<=200; i++)        if (isprime(i)) p[cnt++] = i;    read(n1);read(n2);    for (int i=1; i<=n1+n2; i++){        read(a[i]);read(b[i]);read(c[i]);    }    s = 1;}void build(){    S = 0; T = n1 + n2 + 46 * 46 * 3 + 1;    for (int i=1; i<=n1; i++) ins(S,i,1);    for (int i=n1+1; i<=n1+n2; i++) ins(i,T,1);    int t;    for (int k=1; k<=n1+n2; k++){        x[0] = y[0] = z[0] = 0;        for (int i=0; i<46; i++){            if (a[k] % p[i] == 0) x[++x[0]] = i;            if (b[k] % p[i] == 0) y[++y[0]] = i;            if (c[k] % p[i] == 0) z[++z[0]] = i;        }        t = n1 + n2 + 1;        for (int i=1; i<=x[0]; i++)            for (int j=1; j<=y[0]; j++)                if (k<=n1) ins(k,t+x[i]*46+y[j],INF);                else       ins(t+x[i]*46+y[j],k,INF);        t += 46 * 46;        for (int i=1; i<=x[0]; i++)            for (int j=1; j<=z[0]; j++)                if (k<=n1) ins(k,t+x[i]*46+z[j],INF);                else       ins(t+x[i]*46+z[j],k,INF);        t += 46 * 46;        for (int i=1; i<=y[0]; i++)            for (int j=1; j<=z[0]; j++)                if (k<=n1) ins(k,t+y[i]*46+z[j],INF);                else       ins(t+y[i]*46+z[j],k,INF);    }}bool bfs(){    for (int i=S; i<=T; i++) h[i] = -1, cur[i] = first[i];    int head = 1, tail = 1;    h[q[head] = S] = 1;    for (; head<=tail; head++){        int x = q[head];        for (int t=first[x]; t>0; t=mp[t].next)            if (mp[t].v && h[mp[t].y] == -1)                h[q[++tail] = mp[t].y] = h[x] + 1;    }    return h[T] != -1;}int dfs(int x, int fl){    if (x == T) return fl;    int used = 0, b;    for (int t=cur[x]; t>0; t=cur[x]=mp[t].next){        if (h[mp[t].y] != h[x] + 1) continue;        used += b = dfs(mp[t].y,std::min(fl-used,mp[t].v));        mp[t].v -= b;        mp[t^1].v += b;        if (used == fl) return used;    }    if (!used) h[x] = -1;    return used;}int main(){    init();    build();    while (bfs()) ans += dfs(S,INF);    printf("%d\n",ans);    return 0;}