【SDOI2016】数字配对【建立二分图+费用流求方案数】

题目链接

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首先，我们可以看一下这个推导过程：

如果，

那么，对于，就一定不是质数，一定是它的一个因子。

于是可以看出，这一定是一幅二分图。于是，可以根据二分图的性质来确定了每个点的属于S边还是T边了。

#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 0x3f3f3f3f3f3f3f3f
#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
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 255, maxM = maxN * maxN;
int N, head[maxN], cnt;
ll a[maxN], b[maxN], c[maxN];
struct Eddge
{
    int nex, u, v; ll flow, cost;
    Eddge(int a=-1, int b=0, int c=0, ll d=0, ll f=0):nex(a), u(b), v(c), flow(d), cost(f) {}
}edge[maxM];
inline void addEddge(int u, int v, ll f, ll w)
{
    edge[cnt] = Eddge(head[u], u, v, f, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll f, ll w) { addEddge(u, v, f, w); addEddge(v, u, 0, -w); }
struct MaxFlow_MinCost
{
    int pre[maxN], S, T, node; ll Flow[maxN], dist[maxN];
    queue<int> Q;
    bool inque[maxN];
    inline bool spfa()
    {
        for(int i=0; i<=node; i++) { pre[i] = -1; dist[i] = INF; inque[i] = false; }
        while(!Q.empty()) Q.pop();
        Q.push(S); dist[S] = 0; inque[S] = true; Flow[S] = INF;
        while(!Q.empty())
        {
            int u = Q.front(); Q.pop(); inque[u] = false;
            ll f, w;
            for(int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].v; f = edge[i].flow; w = edge[i].cost;
                if(f && dist[v] > dist[u] + w)
                {
                    dist[v] = dist[u] + w;
                    Flow[v] = min(Flow[u], f);
                    pre[v] = i;
                    if(!inque[v])
                    {
                        inque[v] = true;
                        Q.push(v);
                    }
                }
            }
        }
        return ~pre[T];
    }
    inline ll EK()
    {
        ll ans = 0, res = 0;
        while(spfa() && res >= 0)
        {
            int now = T, las = pre[now];
            while(now ^ S)
            {
                edge[las].flow -= Flow[T];
                edge[las ^ 1].flow += Flow[T];
                now = edge[las].u;
                las = pre[now];
            }
            if(Flow[T] * dist[T] > res) { ans += min(Flow[T], res / dist[T]); break; }
            else { res -= Flow[T] * dist[T]; ans += Flow[T]; }
        }
        return ans;
    }
} MF;
vector<int> vt[maxN];
inline bool Prime(ll x)
{
    for(ll i=2; i * i <= x; i++) if(x % i == 0) return false;
    return true;
}
int col[maxN] = {0};
int que[maxN], top, fail;
void bfs(int S)
{
    col[S] = 1; que[top = fail = 1] = S;
    while(top <= fail)
    {
        int u = que[top ++];
        for(auto v : vt[u])
        {
            if(col[v]) continue;
            col[v] = 3 - col[u];
            que[++fail] = v;
        }
    }
}
inline void init()
{
    cnt = 0;
    for(int i=0; i<=N + 1; i++) head[i] = -1;
    MF.S = 0; MF.T = N + 1;
}
int main()
{
    scanf("%d", &N);
    init();
    for(int i=1; i<=N; i++) scanf("%lld", &a[i]);
    for(int i=1; i<=N; i++) scanf("%lld", &b[i]);
    for(int i=1; i<=N; i++) scanf("%lld", &c[i]);
    for(int i=1; i<=N; i++)
    {
        for(int j=i + 1; j<=N; j++)
        {
            if(a[i] > a[j])
            {
                if(a[i] % a[j]) continue;
                ll tmp = a[i] / a[j];
                if(Prime(tmp)) { vt[i].push_back(j); vt[j].push_back(i); }
            }
            else if(a[i] < a[j])
            {
                if(a[j] % a[i]) continue;
                ll tmp = a[j] / a[i];
                if(Prime(tmp)) { vt[i].push_back(j); vt[j].push_back(i); }
            }
        }
    }
    for(int i=1; i<=N; i++) if(!col[i]) bfs(i);
    for(int i=1; i<=N; i++)
    {
        if(col[i] == 1)
        {
            _add(MF.S, i, b[i], 0);
            for(auto v : vt[i]) _add(i, v, INF, -c[i] * c[v]);
        }
        else
        {
            _add(i, MF.T, b[i], 0);
        }
    }
    MF.node = N + 1;
    printf("%lld\n", MF.EK());
    return 0;
}