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  • 【洛谷P4123】【CQOI2016】—不同的最小割(最小割树)

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    最小割树模板

    #include<bits/stdc++.h>
    using namespace std;
    const int RLEN=1<<22|5;
    inline char gc(){
    	static char ibuf[RLEN],*ib,*ob;
    	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    	return (ib==ob)?EOF:*ib++;
    }
    inline int read(){
    	char ch=gc();
    	int res=0,f=1;
    	while(!isdigit(ch))f^=ch=='-',ch=gc();
    	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    	return f?res:-res;
    }
    #define re register
    #define cs const
    #define ll long long
    #define pb push_back
    #define poly vector<int>
    #define pii pair<int,int>
    #define fi first
    #define se second
    inline void chemx(ll &a,ll b){a<b?a=b:0;}
    inline void chemn(int &a,int b){a>b?a=b:0;}
    cs int N=855,M=8505;
    int n,m,id[N],id1[N];
    map<int,bool> app;
    int tot;
    namespace Dinic{
    	int str,des;
    	int adj[N],nxt[M<<1],to[M<<1],cap[M<<1],cnt=1;
    	inline void addedge(int u,int v,int w){
    		nxt[++cnt]=adj[u],adj[u]=cnt,to[cnt]=v,cap[cnt]=w;
    		nxt[++cnt]=adj[v],adj[v]=cnt,to[cnt]=u,cap[cnt]=w;
    	}
    	int tp[N],lev[N],vis[N],tim;
    	queue<int> q;
    	inline bool bfs(){
    		memset(lev,-1,sizeof(lev));
    		q.push(str),lev[str]=0;
    		while(!q.empty()){
    			int u=q.front();q.pop();
    			for(int e=adj[u];e;e=nxt[e]){
    				int v=to[e];
    				if(cap[e]>0&&lev[v]==-1){
    					lev[v]=lev[u]+1,q.push(v);
    				}
    			}
    		}
    		return lev[des]!=-1;
    	}
    	inline int dfs(int u,int flow){
    		int res=0;
    		if(!flow||u==des)return flow;
    		for(int &e=tp[u];e;e=nxt[e]){
    			int v=to[e];
    			if(lev[v]==lev[u]+1&&cap[e]>0){
    				int now=dfs(v,min(flow-res,cap[e]));
    				cap[e]-=now,cap[e^1]+=now,res+=now;
    				if(flow==res)break;
    			}
    		}
    		return res;
    	}
    	inline int dinic(int s,int t){
    		str=s,des=t;int res=0;
    		while(bfs()){
    			memcpy(tp,adj,sizeof(adj));
    			res+=dfs(str,1e9);
    		}
    		return res;
    	}
    	inline void reset(){
    		for(int i=2;i<=cnt;i+=2)cap[i]=cap[i^1]=(cap[i]+cap[i^1])/2;
    	}
    	void dfs(int u){
    		vis[u]=tim;
    		for(int e=adj[u];e;e=nxt[e]){
    			int v=to[e];
    			if(cap[e]>0&&vis[v]!=tim)dfs(v);
    		}
    	}
    	inline void build(int l,int r){
    		if(l>=r)return;
    		int cnt1=l,cnt2=r;
    		reset();
    		int res=dinic(id[l],id[l+1]);
    		if(!app.count(res))app[res]=1,tot++;
    		tim++,dfs(id[l]);
    		for(int i=l;i<=r;i++)
    		if(vis[id[i]]==tim)id1[cnt1++]=id[i];
    		else id1[cnt2--]=id[i];
    		for(int i=l;i<=r;i++)id[i]=id1[i];
    		build(l,cnt1-1),build(cnt2+1,r);
    	}
    }
    int main(){
    	n=read(),m=read();
    	for(int i=1;i<=m;i++){
    		int u=read(),v=read(),w=read();
    		Dinic::addedge(u,v,w);
    	}
    	for(int i=1;i<=n;i++)id[i]=i;
    	Dinic::build(1,n);cout<<tot;
    }
    
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  • 原文地址:https://www.cnblogs.com/stargazer-cyk/p/12328532.html
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