[CF802C] Heidi and Library Volunteer Recruitment

CF of C title so difficult, the autistic.

Face questions

https://www.luogu.org/problem/CF802C

answer

In addition to the second cut, cost flow problem is generally divided into two kinds:

Thinking in terms of points: column flow conservation equations, the equation seen as a point, as the variable side, there are the typical volunteer recruitment , Delight for a Cat.

Thinking in terms of flows: a representative of a process stream, typically have napkins program issues and the longest interval k can be re-set issues .

model:

Classification model: the smallest cut can be seen as an enhanced version, such as the sequence 48 minute practice.

Adjustment model: Fang Bobo transport coconut , the team returns .

From the perspective of the stream of thinking a good question, but also my first line of minimum cost flow problem.

First, the flow amount of control points is $ K $, representative of the bookcase capacity $ k $. A change of the flow capacity of a representative of the change in the position of the bookshelf.

Each point is split and $ X $ $ x '$, $ 1 $ capacity even with a value of $ $ -INF side, once passed from above, will flow value sharply. Such flows can be distinguished in order to pursue "maximum" or "minimum value" (value stream to see whether the $ <0 $)

Then for $ i, j (i <j) $ connected edge $ (i ', j, 1, c [a [j]]) $, on behalf if the shelf if originally satisfies the first $ I $ a requirement to the first $ j $ to meet the requirement to pay the cost of buying $ j $ (if $ a [i] = a [j] $ then do not pay)

This will be all right.

#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 500
#define S 0
#define T (2*n+1)
#define LL long long
#define ri register int
#define inf 1000000007
using namespace std;

int n,k;
int a[N],c[N];

struct graph {
  vector<int> to,w,c,ed[N];
  LL dis[N]; int cur[N];
  bool vis[N];
  void add_edge(int a,int b,int aw,int ac) {
    to.push_back(b); w.push_back(aw); c.push_back(ac);  ed[a].push_back(to.size()-1);
    to.push_back(a); w.push_back(0);  c.push_back(-ac); ed[b].push_back(to.size()-1);
  }
  bool spfa() {
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    queue<int> q;
    dis[S]=0;q.push(S);vis[S]=1;
    while (!q.empty()) {
      int x=q.front(); q.pop();
      for (ri i=0;i<ed[x].size();i++) {
        int e=ed[x][i];
        if (dis[to[e]]>dis[x]+c[e] && w[e]) {
          dis[to[e]]=dis[x]+c[e];
          if (!vis[to[e]]) vis[to[e]]=1,q.push(to[e]);
        }
      }
      vis[x]=0;
    }
    return dis[T]<inf;
  }
  int dfs(int x,int lim) {
    if (x==T || !lim) return lim;
    LL sum=0; vis[x]=1;
    for (ri &i=cur[x];i<ed[x].size();i++) {
      int e=ed[x][i];
      if (dis[x]+c[e]==dis[to[e]] && w[e] && !vis[to[e]]) {
        int f=dfs(to[e],min(lim,w[e]));
        w[e]-=f; w[1^e]+=f;
        lim-=f; sum+=f;
        if (!lim) return sum;
      }
    }
    return sum;
  }
  LL zkw() {
    LL ret=0;
    while (spfa()) {
      memset(vis,0,sizeof(vis));
      memset(cur,0,sizeof(cur));
      int f=dfs(S,n); k-=f; 
      if (k<0) {
        if ((k+f)*dis[T]<0) ret+=(k+f)*dis[T];
        return ret;
      }
      if (f*dis[T]<0) ret+=f*dis[T];
    }
    return ret;
  }
} G;

int main() {
  scanf("%d %d",&n,&k);
  for (ri i=1;i<=n;i++) scanf("%d",&a[i]);
  for (ri i=1;i<=n;i++) scanf("%d",&c[i]);
  for (ri i=1;i<=n;i++) G.add_edge(S,i,1,c[a[i]]);
  for (ri i=1;i<=n;i++) G.add_edge(i,n+i,1,-inf);
  for (ri i=1;i<=n;i++) G.add_edge(n+i,T,1,0);
  for (ri i=1;i<n;i++) {
    for (ri j=i+1;j<=n;j++) if (a[i]!=a[j]) G.add_edge(n+i,j,1,c[a[j]]); else G.add_edge(n+i,j,1,0);
  }
  cout<<G.zkw()+n*1LL*inf<<endl;
  return 0;
}