nowcoder :Birthday 费用流

传送门

分析

看完之后第一眼反应是个图论问题，想了两种方法，一种是求一个维护联通块个数，一个是网络流建图，后来想想联通块好像不太能写于是往网络流上想
建图比较好建，源点和每一只蜡烛连边，每一只蜡烛和两个点之间连边，流量都为1，费用都为0，剩下的就是费用边应该如何连接的问题了，因为流量和费用并不是一个均衡的关系，所以可以暴力，每个点和汇点之间建50条边，每条边的费用为$k\ast k-(k-1)\ast (k-1)$，然后跑费用流即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N],ne[M],e[M],w[M],f[M],idx;
int n,m,S,T;
int q[N],d[N],pre[N],incf[N];
bool st[N];

void add(int a,int b,int c,int d){
    ne[idx] = h[a],e[idx] = b,f[idx] = c,w[idx] = d,h[a] = idx++;
    ne[idx] = h[b],e[idx] = a,f[idx] = 0,w[idx] = -d,h[b] = idx++;
}

bool spfa(){
    int hh = 0,tt = 1;
    memset(d,0x3f,sizeof d);
    memset(incf,0,sizeof incf);
    q[0] = S,d[S] = 0,incf[S] = INF;
    while(hh != tt){
        int t = q[hh++];
        if(hh == N) hh = 0;
        st[t] = false;
        for(int i = h[t];~i;i = ne[i]){
            int ver = e[i]; 
            if(f[i] && d[ver] > d[t] + w[i]){
                pre[ver] = i;
                d[ver] = d[t] + w[i];
                incf[ver] = min(f[i],incf[t]);
                if(!st[ver]){
                    q[tt++] = ver;
                    if(tt == N) tt = 0;
                    st[ver] = true;
                }
            }
        }
    }
    return incf[T] > 0;
}

void EK(int &flow,int &cost){
    flow = cost = 0;
    while(spfa()){
        int t = incf[T];
        flow += t,cost += t * d[T];
        for(int i = T;i != S;i = e[pre[i] ^ 1]){
            f[pre[i]] -= t;
            f[pre[i] ^ 1] += t;
        }
    }
}

int main() {
    read(n),read(m);
    memset(h,-1,sizeof h);
    S = 0,T = N - 10;
    for(int i = 1;i <= n;i++) add(S,i,1,0);
    for(int i = 1;i <= n;i++){
        int x,y;
        read(x),read(y);
        add(i,x + n,1,0);
        add(i,y + n,1,0);
    }
    for(int i = 1;i <= m;i++)
        for(int j = 1;j <= 50;j++)
            add(i + n,T,1,j * j - (j - 1) * (j - 1));
    int flow,cost;
    EK(flow,cost);
    printf("%d\n",cost);
    return 0;
}