【luoguP3701】「僞模板」主席樹

description

題面太暴力了自己去洛谷看吧

---

analysis

• 我是本來想複習主席樹結果被騙進去複習了一波dinic

• 由題面可以構建網絡流的模型，首先把兩邊法師的數量統計一下，提前先給兩邊的元首加上生命

• 每個人生命 爲$0$就不可以再貢獻，而且對戰成功可以獲得$1$的貢獻

• 那麼源點向$A$方連生命的權值的邊，$B$方向匯點連生命 的權值的邊

• 中間能贏的關係就連$1$邊，最後判斷一下最大流和$m$哪個更小就吼了

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAXN 105
#define MAXM 20005
#define INF 1000000007
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define rep(i,a) for (reg i=last[a];i;i=next[i])

using namespace std;

ll last[MAXM],next[MAXM],tov[MAXM],len[MAXM];
ll depth[MAXN*2],life[MAXN*2];
char a[MAXN*2],s[5];
ll n,m,S,T,tot=1,sum1,sum2;
queue<ll>q;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline void link(ll x,ll y,ll z)
{
	next[++tot]=last[x],last[x]=tot,tov[tot]=y,len[tot]=z;
	next[++tot]=last[y],last[y]=tot,tov[tot]=x,len[tot]=0;
}
inline bool bfs()
{
	while (!q.empty())q.pop();
	memset(depth,0,sizeof(depth));
	q.push(S),depth[S]=1;
	while (!q.empty())
	{
		ll now=q.front();q.pop();
		rep(i,now)if (len[i] && !depth[tov[i]])depth[tov[i]]=depth[now]+1,q.push(tov[i]);
	}
	return depth[T];
}
inline ll dfs(ll x,ll flow)
{
	ll ans=0;
	if (x==T)return flow;
	rep(i,x)if (len[i] && depth[tov[i]]==depth[x]+1)
	{
		ll tmp=dfs(tov[i],min(flow-ans,len[i]));
		if (!tmp)depth[tov[i]]=0;
		ans+=tmp,len[i]-=tmp,len[i^1]+=tmp;
		if (ans==flow)break;
	}
	return ans;
}
inline ll dinic()
{
	ll ans=0;
	while (bfs())
	{
		while (ll x=dfs(S,INF))ans+=x;
	}
	return ans;
}
int main()
{
	freopen("P3701.in","r",stdin);
	n=read(),m=read(),S=2*n+1,T=2*n+2;
	fo(i,1,n)scanf("%s",&s),a[i]=s[0],sum1+=a[i]=='Y';
	scanf("\n");
	fo(i,n+1,n*2)scanf("%s",&s),a[i]=s[0],sum2+=a[i]=='Y';

	fo(i,1,n)life[i]=read()+(a[i]=='J'?sum1:0);
	fo(i,n+1,n*2)life[i]=read()+(a[i]=='J'?sum2:0);

	fo(i,1,n)link(S,i,life[i]);
	fo(i,n+1,n*2)link(i,T,life[i]);
	
	fo(i,1,n)fo(j,n+1,n*2)
	{
		if (a[i]=='J' && (a[j]=='W' || a[j]=='H'))link(i,j,1);
		if (a[i]=='W' && (a[j]=='Y' || a[j]=='E'))link(i,j,1);
		if (a[i]=='H' && (a[j]=='W' || a[j]=='E'))link(i,j,1);
		if (a[i]=='Y' && (a[j]=='H' || a[j]=='J'))link(i,j,1);
		if (a[i]=='E' && (a[j]=='Y' || a[j]=='J'))link(i,j,1);
	}
	printf("%lld\n",min(dinic(),m));
	return 0;
}