【网络流24题】试题库问题

对于每个试题只能属于一个类型的限制,我们从原点向试题连1
然后把每个试题向类别连边,类别向汇点连要求量就可以了
挺板子的说?

/*
@Date    : 2019-08-28 14:20:04
@Author  : Adscn (adscn@qq.com)
@Link    : https://www.cnblogs.com/LLCSBlog
*/
#include<bits/stdc++.h>
using namespace std;
#define IL inline
#define RG register
#define gi getint()
#define gc getchar()
#define File(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
IL int getint()
{
	RG int xi=0;
	RG char ch=gc;
	bool f=0;
	while(ch<'0'||ch>'9')ch=='-'?f=1:f,ch=gc;
	while(ch>='0'&&ch<='9')xi=(xi<<1)+(xi<<3)+ch-48,ch=gc;
	return f?-xi:xi;
}
template<typename T>
IL void pi(T k,char ch=0)
{
	if(k<0)k=-k,putchar('-');
	if(k>=10)pi(k/10,0);
	putchar(k%10+'0');
	if(ch)putchar(ch);
}
const int N=1e3+7;
int S,T;
struct edge{int v,nxt,flow;}e[N*100];
int head[N*2],cur[N*2],cnt;
inline void add(int u,int v,int f){e[++cnt]=(edge){v,head[u],f},head[u]=cnt;}
inline void link(int u,int v,int flow){add(u,v,flow),add(v,u,0);}
int dep[N];
inline bool bfs(){
	memset(dep,-1,sizeof dep);
	static int Q[N<<1];
	int l=1,r=0;
	Q[++r]=S,dep[S]=0;
	while(l<=r)
		for(int p=Q[l++],i=head[p],v;~i;i=e[i].nxt)
			if(e[i].flow&&dep[v=e[i].v]==-1)
				dep[v]=dep[p]+1,Q[++r]=v;
	return ~dep[T];
}
inline int dfs(int p,int restflow){ 
	if(p==T||!restflow)return restflow;
	int sumflow=0;
	for(int &i=cur[p],flow,v;~i;i=e[i].nxt)
		if(e[i].flow&&dep[v=e[i].v]==dep[p]+1&&(flow=dfs(v,min(e[i].flow,restflow))))
		{
			restflow-=flow,sumflow+=flow;
			e[i].flow-=flow,e[i^1].flow+=flow;
			if(!restflow)break;
		}
	return sumflow;
}
inline int dinic(void){
	int maxflow=0;
	while(bfs())memcpy(cur,head,sizeof head),maxflow+=dfs(S,INT_MAX);
	return maxflow;
}
inline void init(){memset(head,cnt=-1,sizeof head);}
queue<int>Output[50];
int main(void)
{
	int k=gi,n=gi;
	init();
	S=k+n+1,T=k+n+2;
	int m=0;
	for(int i=1;i<=k;++i)
	{
		int t=gi;
		m+=t;
		link(n+i,T,t);
	}
	for(int i=1;i<=n;++i){
		int num=gi;
		while(num--)link(i,n+gi,1);
	}
	for(int i=1;i<=n;++i)link(S,i,1);
	int mx=dinic();
	if(mx!=m)return printf("No Solution!"),0;
	for(int i=1;i<=n;++i)
		for(int j=head[i];~j;j=e[j].nxt)
			if((j&1)==0&&!e[j].flow)
			{	
				Output[e[j].v-n].push(i);
				break;
			}
	for(int i=1;i<=k;++i){
		pi(i,':');
		while(!Output[i].empty())pi(Output[i].front(),' '),Output[i].pop();
		puts("");
	}
	return 0;
}
原文地址:https://www.cnblogs.com/LLCSBlog/p/11424922.html