题意:用最小的点来覆盖全部的边,因为二分图里面最大的匹配就是最小覆盖,所以直接匹配一下即可
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#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std; const int MAXN = 1505;
const int oo = 1e9+7; struct Edge{int v, next;}e[MAXN*4];///数组开小RE了一次
int Head[MAXN], cnt; int Mx[MAXN], My[MAXN];
int dx[MAXN], dy[MAXN];
int used[MAXN], N, depth; void InIt()
{
cnt = 0;
memset(Head, -1, sizeof(Head));
memset(Mx, -1, sizeof(Mx));
memset(My, -1, sizeof(My));
}
void AddEdge(int u, int v)
{
e[cnt].v = v;
e[cnt].next = Head[u];
Head[u] = cnt++;
}
bool BFS()
{
queue<int> Q;
depth = oo; memset(dx, false, sizeof(dx));
memset(dy, false, sizeof(dy)); for(int i=0; i<N; i++)
{
if( Mx[i] == -1 )
{
dx[i] = true;
Q.push(i);
}
} while(Q.size())
{
int u = Q.front();Q.pop(); if(dx[u] > depth)break; for(int j=Head[u]; j!=-1; j=e[j].next)
{
int v = e[j].v;
if( dy[v] == false )
{
dy[v] = dx[u] + 1;
if(My[v] == -1)
depth = dy[v];
else
{
dx[ My[v] ] = dy[v] + 1;
Q.push( My[v] );
}
}
}
} return depth != oo;
}
bool DFS(int i)
{
for(int j=Head[i]; j!=-1; j=e[j].next)
{
int v = e[j].v;
if( used[v] == false && dx[i] == dy[v]-1 )
{
used[v] = true;
if(My[v] != -1 && dy[v] == depth)
continue;
if(My[v] == -1 || DFS(My[v]))
{
My[v] = i;
Mx[i] = v; return true;
}
}
} return false;
}
int Karp()
{
int ans = 0; while( BFS() == true)
{
memset(used, false, sizeof(used));
for(int i=0; i<N; i++)
{
if( Mx[i] == -1 && DFS(i) )
ans ++;
}
} return ans;
} int main()
{
while(scanf("%d", &N) != EOF)
{
int M, i, u, v; InIt(); for(i=0; i<N; i++)
{
scanf("%d:(%d)", &u, &M);
while(M--)
{
scanf("%d", &v);
AddEdge(u, v);
AddEdge(v, u);
}
} int ans = Karp(); printf("%d\n", ans/2);
} return 0;
}