【 bzoj 2661 】 [BeiJing wc2012]连连看 - 拆点费用流

　　每个数 i 拆成两个点 Li 和 Ri ，连边 S→Li,Ri→T 流量为1费用为0，每对有关系的数 (x,y) 连边 Lx→Ry,Ly→Rx 流量为1费用为 x+y ，打个表可以发现边数不会很多，直接跑最大费用最大流。
　　

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for (int i = a , _ = b ; i <= _ ; i ++)
#define per(i,a,b) for (int i = a , _ = b ; i >= _ ; i --)
#define fore(i,u) for (int i = head[u] ; i ; i = nxt[i])

inline int rd() {
    char c = getchar();
    while (!isdigit(c)) c = getchar() ; int x = c - '0';
    while (isdigit(c = getchar())) x = x * 10 + c - '0';
    return x;
}

const int maxn = 2017;
const int maxm = 500007;
const int inf = 1000000000;

typedef int arr[maxn];
typedef int adj[maxm];

arr head , dis , pre , vis , lid , rid;
adj fr , to , cap , cost , nxt , flow;

queue<int> Q;

inline int gcd(int a , int b) { return b ? gcd(b , a % b) : a; }

inline int sqr(int x) { return x * x; }

inline int _sqrt(int x) {
    int t = (int) sqrt(x + 0.5);
    if (t * t != x) return -1;
    return t;
}

int A , B , ett , S , T , tot;
int mx_cost , mx_flow;

void input() {
    ett = 1;
    A = rd() , B = rd();
}

inline void ins(int u , int v , int c , int w) {
    fr[++ ett] = u , to[ett] = v , cap[ett] = c , cost[ett] = w , nxt[ett] = head[u] , head[u] = ett;
    fr[++ ett] = v , to[ett] = u , cap[ett] = 0 , cost[ett] = -w, nxt[ett] = head[v] , head[v] = ett;
}

bool spfa() {
    rep (i , 1 , tot) dis[i] = -inf;
    dis[S] = 0 , vis[S] = 1 , pre[S] = 0;
    Q.push(S);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        vis[u] = 0;
        fore (i , u) if (cap[i] > flow[i]) {
            int v = to[i];
            if (dis[v] < dis[u] + cost[i]) {
                dis[v] = dis[u] + cost[i];
                pre[v] = i;
                if (!vis[v]) Q.push(v) , vis[v] = 1;
            }
        }
    }
    if (dis[T] == -inf) return 0;
// assert(!pre[S]);
    int a = inf;
    for (int e = pre[T];e;e = pre[fr[e]]) a = min(a , cap[e] - flow[e]);
    mx_cost += dis[T] * a , mx_flow += a;
    for (int e = pre[T];e;e = pre[fr[e]])
        flow[e] += a , flow[e ^ 1] -= a;
    return 1;
}

#define L(x) lid[x]
#define R(x) rid[x]

//vector<int> lk[maxn];

void solve() {
    rep (i , A , B) lid[i] = ++ tot , rid[i] = ++ tot;
    rep (x , A , B) rep (y , A , x - 1) {
        int z = _sqrt(sqr(x) - sqr(y));
        if (z == -1 || gcd(z , y) != 1) continue;
// lk[x].push_back(ett + 1);
        ins(L(x) , R(y) , 1 , x + y);
// lk[x].push_back(ett + 1);
        ins(L(y) , R(x) , 1 , x + y);
    }
    S = ++ tot , T = ++ tot;
    rep (i , A , B) ins(S , L(i) , 1 , 0) , ins(R(i) , T , 1 , 0);
    for (;spfa();)/* printf("%d %d\n" , mx_flow , mx_cost)*/;
/* rep (x , A , B) for (int i = 0;i < lk[x].size();i ++) if (cap[lk[x][i]] == flow[lk[x][i]]) printf("%d %d\n" , x , cost[lk[x][i]] - x);*/
    printf("%d %d\n" , mx_flow / 2 , mx_cost / 2);
}

int main() {
    #ifndef ONLINE_JUDGE
        freopen("data.txt" , "r" , stdin);
    #endif
    input();
    solve();
    return 0;
}