P3701 主主树(最大流)

P3701 主主树(最大流)

S-T很显然，然后寿命作为权 连S-T。

然后那个续命，注意自己不会掉血，所以等价于J的生命加上同阵营的YYY的人数。

然后根据克制关系建权1的边，然后跑最大流与m取最小值。

// Problem: P3701 主主树
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3701
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// Date: 2022-05-31 09:42:56
// --------by Herio--------

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=2e3+5,M=2e5+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = {402653189,805306457,1610612741,998244353};
#define mst(a,b) memset(a,b,sizeof a)
#define db double
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n){
  for(int i=1;i<n;i++)
    printf("%d ",a[i]);
  printf("%d\n",a[n]); 
}
template <typename T>   //x=max(x,y)  x=min(x,y)
void cmx(T &x,T y){
  if(x<y) x=y;
}
template <typename T>
void cmn(T &x,T y){
  if(x>y) x=y;
}
struct Dinic{
//Dinic O(n^2m)
  int n,m,st,ed;
  int id(int x,int y){
    return (x-1)*m+y;
  }
  struct edge{
    int to,nt;
    ll w;
  }e[M];
  int h[N],cur[N],cnt,dep[N];
  void init(int _st,int _ed){
    st=_st,ed=_ed;
    cnt=1;mst(h,0);
  }
  Dinic(int _st=0,int _ed=0){init(_st,_ed);}
  void add(int u,int v,ll w){
    e[++cnt]={v,h[u],w},h[u]=cnt;
    e[++cnt]={u,h[v],0},h[v]=cnt;
  }
  map<string,int>mp;
  int bl[N];
  void solve(){
    mp["J"]=1;
    mp["HK"]=2;
    mp["W"]=3;
    mp["YYY"]=4;
    mp["E"]=5;
    cnt = 1,mst(h,0);
    scanf("%d%d",&n,&m);
    st = 0,ed = n+n+1;
    int sum = 0,sum1=0;
    rep(i,1,n){
      string s;cin>>s;
      bl[i] = mp[s];
      if(bl[i] == 4) sum++;
    }
    rep(i,1,n){
      string s;cin>>s;
      bl[i+n] = mp[s];
      if(bl[i+n] == 4) sum1++;
    }

    rep(i,1,n){
      int x;cin>>x;
      if(bl[i]==1)
        x+=sum;
      //printf("(%d,%d,%d)\n",st,i,x);
      add(st,i,x);
    }
    rep(i,1,n){
      int x;cin>>x;
      if(bl[i+n] == 1) x+=sum1;
      //("(%d,%d,%d)\n",i+n,ed,x);
      add(i+n,ed,x);
    }
    //("%d,%d\n",sum,sum1);
    rep(i,1,n)
      rep(j,1,n){
        int ok  =0;
        if(bl[i]==1 && (bl[j+n]==2 || bl[j+n]==3))
          ok = 1;
        else if(bl[i]==2 && (bl[j+n]==3 || bl[j+n]==5))
          ok = 1;
        else if(bl[i]==3 && (bl[j+n]==4 || bl[j+n]==5))
          ok = 1;
        else if(bl[i]==4 && (bl[j+n]==1 || bl[j+n]==2))
          ok = 1;
        else if(bl[i]==5 && (bl[j+n]==1 || bl[j+n]==4))
          ok = 1;
        if(ok)
        add(i,j+n,1);//,printf("(%d,%d)\n",i,j+n);
        
      }
    printf("%lld\n",min(m+0LL,dinic()));
  }
  ll dfs(int u,ll c){ //search for augment path
    if(u==ed) return c;
    ll res=c;
    for(int &i=cur[u];i;i=e[i].nt){
      int v=e[i].to; ll w=e[i].w;
      if(w&&dep[v]==dep[u]+1){
        ll now=dfs(v,min(res,w));
        if(!now) dep[v]=1;
        else e[i].w-=now,e[i^1].w+=now,res-=now;
      }
      if(!res) return c;
    }return c-res;
  } 
  bool bfs(){   //layer the graph
    queue<int>q;q.push(st);mst(dep,0),dep[st]=1;
    while(!q.empty()){
      int u=q.front();q.pop();cur[u]=h[u];
      for(int i=h[u];i;i=e[i].nt){
        int v=e[i].to;ll w=e[i].w;
        if(w&&!dep[v]) dep[v]=dep[u]+1,q.push(v);
      }
    }return dep[ed];
  } 
  ll dinic(){
    ll s=0;
    while(bfs()) s+=dfs(st,inf);
    return s;
  }
}G;
int main(){
  G.solve();
  return 0;
}