这题很简单啊
从源点向k类题目分别连流量为所需数量的边
从每道题向汇点连一条流量为1的边（每题只能用1次）
从类型向对应的题目连一条流量为1的边
跑一遍最大流
如果最大流小于所需题目数量，就无解
否则对于每个类型，看每道题是否要选
#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define M 80005
#define N 2005
//#define getchar nc
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
register int x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*f;
}
inline void write(register int x)
{
if(!x)putchar('0');if(x<0)x=-x,putchar('-');
static int sta[20];register int tot=0;
while(x)sta[tot++]=x%10,x/=10;
while(tot)putchar(sta[--tot]+48);
}
inline int Min(register int a,register int b)
{
return a<b?a:b;
}
struct node{
int to,nxt,v;
}e[M];
int head[N],cnt=1;
inline void add(register int u,register int v,register int val)
{
e[++cnt]=(node){v,head[u],val};
head[u]=cnt;
}
int n,k,s,t,nn,maxflow=0,sum=0;
int dep[N],gap[N],cur[N];
inline void bfs()
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
dep[t]=0;
++gap[dep[t]];
queue<int> q;
q.push(t);
while(!q.empty())
{
int u=q.front();
q.pop();
for(register int i=head[u];i;i=e[i].nxt)
{
int v=e[i].to;
if(dep[v]!=-1)
continue;
q.push(v);
dep[v]=dep[u]+1;
++gap[dep[v]];
}
}
}
inline int dfs(register int u,register int flow)
{
if(u==t)
{
maxflow+=flow;
return flow;
}
int used=0;
for(register int i=cur[u];i;i=e[i].nxt)
{
cur[u]=i;
int v=e[i].to;
if(e[i].v&&dep[v]+1==dep[u])
{
int tmp=dfs(v,Min(e[i].v,flow-used));
if(tmp)
{
e[i].v-=tmp;
e[i^1].v+=tmp;
used+=tmp;
}
if(used==flow)
return used;
}
}
--gap[dep[u]++]==0?dep[s]=nn+1:++gap[dep[u]];
return used;
}
inline void ISAP()
{
bfs();
while(dep[s]<nn)
{
memcpy(cur,head,sizeof(head));
dfs(s,inf);
}
}
int main()
{
k=read(),n=read();
s=0,t=k+n+1;
for(register int i=1;i<=k;++i)
{
int x=read();
add(s,i,x),add(i,s,0),sum+=x;
}
for(register int i=1;i<=n;++i)
{
int p=read();
for(register int j=1;j<=p;++j)
{
int x=read();
add(x,i+k,1),add(i+k,x,0);
}
add(i+k,t,1),add(t,i+k,0);
}
nn=n+k+2;
ISAP();
if(maxflow==sum)
for(register int u=1;u<=k;++u)
{
write(u),putchar(':'),putchar(' ');
for(register int i=head[u];i;i=e[i].nxt)
if(e[i].to!=0&&e[i].v==0)
write(e[i].to-k),putchar(' ');
puts("");
}
else
puts("No Solution!");
return 0;
}