[BJOI2012] 连连看 - 费用流

Description

给出一个闭区间 \([a,b]\) 中的全部整数，如果其中某两个数 \(x\)，\(y\)（\(x > y\)）的平方差 \(x^2-y^2\) 是一个完全平方数 \(z^2\)，并且 \(y\) 与 \(z\) 互质，那么就可以将 \(x\) 和 \(y\) 连起来并且将它们一起消除，同时得到 \(x+y\) 点分数。消除的数对尽可能多的前提下，得到足够的分数。\(a,b \le 1000\)

Solution

关键是要通过正反建边来保证数字不会被重复使用，最终结果除以 2 即可

#include <bits/stdc++.h>
using namespace std;
#define int long long
// Init: init() !!!!!
// Input: make(u,v,cap,cost)
// Solver: solve(s,t)
// Output: ans, cost
namespace flow {
const int N = 100005;
const int M = 1000005;
const int inf = 1e+12;
struct Edge {
    int p, c, w, nxt = -1;
} e[N];
int s, t, tans, ans, cost, ind, bus[N], qhead = 0, qtail = -1, qu[M],vis[N], dist[N];

void graph_link(int p, int q, int c, int w) {
    e[ind].p = q;
    e[ind].c = c;
    e[ind].w = w;
    e[ind].nxt = bus[p];
    bus[p] = ind;
    ++ind;
}
void make(int p, int q, int c, int w) {
    graph_link(p, q, c, w);
    graph_link(q, p, 0, -w);
}
int dinic_spfa() {
    qhead = 0;
    qtail = -1;
    memset(vis, 0x00, sizeof vis);
    memset(dist, 0x3f, sizeof dist);
    vis[s] = 1;
    dist[s] = 0;
    qu[++qtail] = s;
    while (qtail >= qhead) {
        int p = qu[qhead++];
        vis[p] = 0;
        for (int i = bus[p]; i != -1; i = e[i].nxt)
            if (dist[e[i].p] > dist[p] + e[i].w && e[i].c > 0) {
                dist[e[i].p] = dist[p] + e[i].w;
                if (vis[e[i].p] == 0)
                    vis[e[i].p] = 1, qu[++qtail] = e[i].p;
            }
    }
    return dist[t] < inf;
}
int dinic_dfs(int p, int lim) {
    if (p == t)
        return lim;
    vis[p] = 1;
    int ret = 0;
    for (int i = bus[p]; i != -1; i = e[i].nxt) {
        int q = e[i].p;
        if (e[i].c > 0 && dist[q] == dist[p] + e[i].w && vis[q] == 0) {
            int res = dinic_dfs(q, min(lim, e[i].c));
            cost += res * e[i].w;
            e[i].c -= res;
            e[i ^ 1].c += res;
            ret += res;
            lim -= res;
            if (lim == 0)
                break;
        }
    }
    return ret;
}
void solve(int _s,int _t) {
    s=_s; t=_t;
    while (dinic_spfa()) {
        memset(vis, 0x00, sizeof vis);
        ans += dinic_dfs(s, inf);
    }
}
void init() {
    memset(bus, 0xff, sizeof bus);
}
}



signed main() {
    int a,b;
    cin>>a>>b;
    flow::init();
    for(int i=a;i<=b;i++) {
        for(int j=a;j<=b;j++) {
            int x=i*i-j*j;
            if(i<=j) continue;
            int z=((int)sqrt(x));
            if(__gcd(z,j)!=1) continue;
            if(z*z!=x) continue;
            flow::make(i,j+b,1,-i-j);
            flow::make(j,i+b,1,-i-j);
        }
        flow::make(2*b+1,i,1,0);
        flow::make(i+b,2*b+2,1,0);
    }
    flow::solve(2*b+1,2*b+2);
    cout<<flow::ans/2<<" "<<-flow::cost/2;
}

原文地址：https://www.cnblogs.com/mollnn/p/12883258.html