[题解]UVA1194 Machine Schedule

传送门

题目大意：两台机器，分别有 (m) 个模式，初始时处于关机状态。有(k) 个任务，第 (i) 个任务可以在机器 (A) 上用 (a_i) 模式完成，也可以在机器 (B) 上用 (b_i) 模式完成。一台机器每更换一次模式需要关机一次。求完成所有任务所需的关机次数最小值。嘤语不好，轻喷

考虑网络流。可以由原点向机器 (A) 的 (n) 种模式连边，由机器 (B) 的 (m) 种模式向会点连边。对于每一个任务，由 (a_i) 向 (b_i) 连边。

跑一遍 (dinic) 即可

代码

#include<bits/stdc++.h>
namespace IO
{
	using namespace std;
	char buf[1<<22],Out[1<<22],*p1=buf,*p2=buf;
	int p3=-1,f=0;
	inline void file()
	{
		#ifdef NTFOrz
		freopen("file.in","r",stdin);
		#endif
	}
	inline char getchar()
	{
		return (p1==p2)&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;
	}
	inline void flush(){fwrite(Out,1,p3+1,stdout),p3=-1;}
	inline void check(){if (p3>(1<<21)) flush();}
	inline void putchar(char a){Out[++p3]=a;check();}
	inline void putline(string s)
	{
		int tmp=p3;
		while (p3-tmp<s.size()) putchar(s[p3-tmp]);
		putchar('
');
	}
	inline char gc()
	{
		char c=getchar();
		while (!c||c=='
'||c==' ') c=getchar();
		return c;
	}
}
namespace my_std
{
	using namespace std;
	#define int long long
	#pragma GCC optimize(2)
	#pragma GCC optimize(3) 
	#pragma GCC optimize("Ofast")
	#define getchar IO::getchar
	#define putchar IO::putchar
    #define Re register
    #define rep(i,a,b) for (Re int i=(a);i<=(b);i++)
    #define drep(i,a,b) for (Re int i=(a);i>=(b);i--)
    #define repv(i,v) if (v.size()) rep(i,0,v.size()-1)
    #define drepv(i,v) if (v.size()) drep(i,v.size()-1,0)
    #define go(u) for (Re int i=head[(u)];i;i=e[i].nxt)
    #define pf printf
    #define writesp(x) write(x),putchar(' '),IO::check()
    #define writeln(x) write(x),putchar('
'),IO::check()
    #define mem(x,v) memset(x,v,sizeof(x))
    #define pb push_back
    #define NTF return
    #define AKIOI 0
    #define fir first
    #define sec second
    #define MP make_pair
    #define all(x) x.begin(),x.end()
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    typedef complex<double> Complex;
    const int INF=0x7fffffff;
    inline int read()
    {
        int sum=0,f=0;char c=getchar();
        while (!isdigit(c)) f|=(c=='-'),c=getchar();
        while (isdigit(c)) sum=(sum<<1)+(sum<<3)+(c^48),c=getchar();
        return f?-sum:sum;
    }
    void write(int k)
    {
        if (k<0) putchar('-'),k=-k;
        if (k>=10) write(k/10);
        putchar(k%10+48);
    }
	#define templ template<typename T>
    templ inline void chkmax(T &x,T y){if (x<y) x=y;}
    templ inline void chkmin(T &x,T y){if (x>y) x=y;}
}
using namespace my_std;
const int N=510;
int n,m,k,cnt=1,dep[N<<1],head[N<<1],S,T;
struct edge
{
	int to,nxt,flow;
}e[N*N];
inline void add(int u,int v,int flow){e[++cnt].to=v,e[cnt].nxt=head[u],head[u]=cnt,e[cnt].flow=flow;}
inline void ins(int u,int v,int flow){add(u,v,flow),add(v,u,0);}
inline bool bfs()
{
	static queue<int>q;
	mem(dep,0),dep[S]=1,q.push(S);
	while (!q.empty())
	{
		int u=q.front();
		go(u)
		{
			int v=e[i].to;
			if (e[i].flow&&!dep[v])
			{
				dep[v]=dep[u]+1;
				q.push(v);
			}
		}
		q.pop();
	}
	return dep[T];
}
int dfs(int u,int in)
{
	if (u==T) return in;
	int out=0;
	go(u)
	{
		int v=e[i].to;
		if (e[i].flow&&dep[v]==dep[u]+1)
		{
			int res=dfs(v,min(in,e[i].flow));
			in-=res,out+=res,e[i].flow-=res,e[i^1].flow+=res;
		}
	}
	if (out==0) dep[u]=0;
	return out;
}
inline int dinic(){int ans=0;while (bfs()) ans+=dfs(S,INF);return ans;}
inline void clear(){cnt=1,mem(head,0);}
inline void solve()
{
	m=read(),k=read();
	S=0,T=n+m+1;
	rep(i,1,n) ins(S,i,1);
	rep(i,1,m) ins(i+n,T,1);
	rep(i,1,k)
	{
		int TAT=read(),u=read(),v=read()+n;
		if (u*v==0) continue;
		ins(u,v,INF);
	}
	writeln(dinic());
	clear();
}
signed main()
{
	IO::file();
	n=read();
	while (n) solve(),n=read();
	IO::flush();
	return 0;
}