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题意:给定一颗无向树,求最小点覆盖。
连通的有向图的情况,将点集copy到L和R集合,建一个二分图,由L集合的点向R集合连一条有向边,这该二分图的最大匹配数就等于最小点覆盖数。连边如(0, 1),则只连L0->R1
本题无向图的连边如(0, 1),则要连L0->R1和L1->R0,最小点覆盖数==最大匹配数/2
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1510;
const int M = N*N;
int linker[N];
bool vis[N];
int to[M],nxt[M];
int head[N],tot;
int n;
void addedge(int u, int v)
{
++tot;
to[tot] = v;
nxt[tot] = head[u];
head[u] = tot;
}
bool dfs(int u)
{
for(int i = head[u]; ~i; i = nxt[i])
{
int v = to[i];
if(!vis[v])
{
vis[v] = 1;
if(linker[v]==-1 || dfs(linker[v]))
{
linker[v] = u;
return 1;
}
}
}
return 0;
}
int hungary()
{
int ret = 0;
memset(linker, -1, sizeof(linker));
for(int i = 0; i < n; ++i)
{
memset(vis, 0, sizeof(vis));
if(dfs(i))
++ret;
}
return ret;
}
int main()
{
while(~scanf("%d", &n))
{
memset(head, -1, sizeof(head));
tot = -1;
for(int i = 1; i <= n; ++i)
{
int u, v, k;
scanf("%d:(%d)", &u, &k);
while(k--)
{
scanf("%d", &v);
addedge(u, v);
addedge(v, u);
}
}
printf("%d\n", hungary()>>1);
}
return 0;
}