CF498C. Array and Operations [二分图]

CF498C. Array and Operations

题意:

给定一个长为 n 的数组,以及 m 对下标 (a, b) 且满足 a + b 为奇数,每次操作可以将同一组的两个数同时除以一个公约数

问最多能进行多少次操作

\[1≤n,m ≤100,1≤ai ≤10^9\]


根据奇偶性二分图定理此题必定考二分图

贪心,每次除一个质数

质数之间是独立的,可以分开考虑每一个质因子

建图:s -x中质因子p数量-> x -inf-> y -y中质因子p数量-> t

最大权匹配就是这个质因子能带来的最多操作数

注意质因子分解别写错了,最后判x>1

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <set>
#include <map>

#define fir first
#define sec second
using namespace std;
const int N = 105, inf = 1e9;


int n, m, a[N];
pair<int, int> b[N];
map<int, int> li[N];
set<int> s;

void fac(int p) { 
    int x = a[p], sx = sqrt(x) + 1; //printf("fac %d\n", a[p]);
    for(int i=2; i<=sx; i++) if(x % i == 0) { //printf("iii %d\n", i);
        int cnt = 0;
        while(x%i == 0) cnt++, x/=i; //printf("x %d\n", x);
        //li[p].push_back(make_pair(i, cnt));
        li[p][i] = cnt;
        s.insert(i);
    }
    if(x > 1) li[p][x] = 1, s.insert(x);
}

namespace mf {
    int s, t;
    struct edge {int v, ne, c, f;} e[1005];
    int cnt=1, h[N];
    void ins(int u, int v, int c) { //printf("ins %d %d %d\n", u, v, c);
        e[++cnt] = (edge) {v, h[u], c, 0}; h[u] = cnt;
        e[++cnt] = (edge) {u, h[v], 0, 0}; h[v] = cnt;
    }
    int cur[N], vis[N], d[N], head, tail, q[N];
    bool bfs() {
        memset(vis, 0, sizeof(vis));
        head = tail = 1;
        q[tail++] = s; d[s] = 0; vis[s] = 1;
        while(head != tail) {
            int u = q[head++];
            for(int i=h[u]; i; i=e[i].ne) {
                int v = e[i].v;
                if(!vis[v] && e[i].c > e[i].f) {
                    vis[v] = 1;
                    d[v] = d[u] + 1;
                    q[tail++] = v;
                    if(v == t) return true;
                }
            }
        }
        return false;
    }
    int dfs(int u, int a) {
        if(u==t || a==0) return a;
        int flow = 0, f;
        for(int &i=cur[u]; i; i=e[i].ne) {
            int v = e[i].v;
            if(d[v] == d[u]+1 && (f = dfs(v, min(a, e[i].c-e[i].f))) > 0) {
                flow += f;
                e[i].f += f;
                e[i^1].f -= f;
                a -= f;
                if(a==0) break;
            }
        }
        if(a) d[u] = -1;
        return flow;
    }

    void build(int p) {
        cnt = 1;
        memset(h, 0, sizeof(h));
        s=0; t=n+1;
        for(int i=1; i<=m; i++) ins(b[i].fir, b[i].sec, inf);
        for(int i=1; i<=n; i++) if(li[i].count(p)) {
            if(i & 1) ins(s, i, li[i][p]);
            else ins(i, t, li[i][p]);
        }
    }

    int solve(int p) { //printf("\nsolve %d\n", p);
        build(p);
        int flow = 0;
        while(bfs()) {
            for(int i=s; i<=t; i++) cur[i] = h[i];
            flow += dfs(s, inf);
        }
        return flow;
    }

}
int main() {
    //freopen("in", "r", stdin);
    ios::sync_with_stdio(false); cin.tie(); cout.tie();

    cin >> n >> m;
    for(int i=1; i<=n; i++) cin >> a[i];
    for(int i=1; i<=m; i++) {
        cin >> b[i].fir >> b[i].sec;
        if(b[i].sec & 1) swap(b[i].fir, b[i].sec);
    }

    for(int i=1; i<=n; i++) fac(i);
    int ans = 0;
    
    for(set<int>::iterator it = s.begin(); it != s.end(); it++) ans += mf::solve(*it);
    cout << ans;
}
posted @ 2018-07-17 19:24 Candy? 阅读(...) 评论(...) 编辑 收藏