最小不可相交路径覆盖。

点数-最大匹配数

代码如下：

#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int max_n = 2000;
const int inf = 2e9;
struct edge{
    int to,next;
}E[max_n<<1];
int head[max_n];
int cnt=1;
void add(int from,int to){
    E[cnt].to=to;E[cnt].next=head[from];
    head[from]=cnt++;
}
int leftTo[max_n], rightTo[max_n];
int distleft[max_n], distright[max_n];
int dist;
bool seacherpath(int left_tot, int right_tot) {
    fill(distleft, distleft + left_tot + 1, -1);
    fill(distright, distright + right_tot + 1, -1);
    queue<int> que;dist = inf;
    for (int i = 1;i <= left_tot;++i)
        if (leftTo[i] == -1)
            que.push(i);
    while (!que.empty()) {
        int u = que.front();que.pop();
        if (distleft[u] > dist)break;
        for (int i = head[u];i;i = E[i].next) {
            int v = E[i].to;
            if (distright[v] != -1)continue;
            distright[v] = distleft[u] + 1;
            if (rightTo[v] == -1)dist = distright[v];
            else {
                distleft[rightTo[v]] = distright[v] + 1;
                que.push(rightTo[v]);
            }
        }
    }return dist != inf;
}
bool matchpath(int u) {
    for (int i = head[u];i;i = E[i].next) {
        int v = E[i].to;
        if (distright[v] != distleft[u] + 1)continue;
        if (distright[v] == dist && rightTo[v] != -1)continue;
        distright[v] = -1;
        if (rightTo[v] == -1 || matchpath(rightTo[v])) {
            leftTo[u] = v;
            rightTo[v] = u;
            return true;
        }
    }return false;
}
int HK(int left_tot, int right_tot) {
    fill(leftTo, leftTo + left_tot + 1, -1);
    fill(rightTo, rightTo + right_tot + 1, -1);
    int ans = 0;
    while (seacherpath(left_tot, right_tot)) {
        for (int i = 1;i <= left_tot;++i)
            if (leftTo[i] == -1 && matchpath(i))
                ++ans;
    }return ans;
}
int n;
int main(){
    while(scanf("%d",&n)!=EOF){
        fill(head,head+n+10,0);
                cnt=1;
        for (int i=1;i<=n;++i){
            int u,nu;
            scanf("%d:(%d)",&u,&nu);
            ++u;
            while (nu--){
                int v;scanf("%d",&v);
                add(u,++v);add(v,u);
            }
        }
        printf("%d\n",HK(n,n)/2);
    }
}