题目链接：点击打开链接

题意：

给定n个人，m个终点

下面n行表示每个人可以去m个点。

每个人只能去一个点。

输出任意一个方案使得每个点至少有2个人到达。

若存在输出m行，第一个数字表示i这个点来了几个人，后面是人的点标。

思路：

建一个二部图n-m，然后m到汇点限流2，判断是否满流，若不满流就无解。

若满流则其他的人随便走。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <set>
#include <algorithm>
using namespace std;
typedef int ll;
#define N 420
#define M 100000
#define inf 107374182
struct Edge{
    ll from, to, cap, nex;
}edge[M*2];
ll head[N], edgenum;
void add(ll u, ll v, ll cap, ll rw = 0){
    Edge E = {u, v, cap, head[u]};
    edge[edgenum] = E;
    head[u] = edgenum ++;
    Edge E2 = {v, u, rw, head[v]};
    edge[edgenum] = E2;
    head[v] = edgenum ++;
}
ll sign[N];
bool BFS(ll from, ll to){
    memset(sign, -1, sizeof sign);
    sign[from] = 0;
    queue<ll>q;
    q.push(from);
    while( !q.empty() ) {
        ll u = q.front(); q.pop();
        for(ll i = head[u]; ~i; i = edge[i].nex)
        {
            ll v = edge[i].to;
            if(sign[v] == -1 && edge[i].cap){
                sign[v] = sign[u] +1, q.push(v);
                if(sign[to] != -1) return true;
            }
        }
    }
    return false;
}
ll Stack[N], top, cur[N];
ll Dinic(ll from, ll to){
    ll ans = 0;
    while( BFS(from, to) )
    {
        memcpy(cur, head, sizeof head);
        ll u = from; top = 0;
        while(1)
        {
            if(u==to)
            {
                ll flow = inf, loc;
                for(ll i = 0; i < top; i++)
                    if(flow > edge[Stack[i]].cap)
                    {
                        flow = edge[Stack[i]].cap;
                        loc = i;
                    }
                for(ll i = 0; i < top; i++)
                {
                    edge[ Stack[i] ].cap -= flow;
                    edge[Stack[i]^1].cap += flow;
                }
                ans += flow;
                top = loc;
                u = edge[Stack[top]].from;
            }
            for(ll i = cur[u]; ~i; cur[u] = i = edge[i].nex)
                if(edge[i].cap && (sign[u]+1 == sign[edge[i].to]))break;
            if(cur[u] != -1){
                Stack[top++] = cur[u];
                u = edge[cur[u]].to;
            }
            else
            {
                if(top==0)break;
                sign[u] = -1;
                u = edge[Stack[--top]].from;
            }
        }
    }
    return ans;
}
void init(){memset(head, -1, sizeof head); edgenum = 0;}
int n, m, from, to, to2, ans[N], G[N];
vector<int>D[N];
bool solve(){
    from = 0; to = n+m+1; to2 = to+1;
    init();
    for(int i = 1; i <= n; i++)
    {
        G[i] = -1;
        add(from, i, 1);
        int x, y; scanf("%d",&x);
        while(x--){
            scanf("%d",&y);
            G[i] = y;
            add(i, n+y, 1);
        }
    }
    for(int i = 1; i <= m; i++)
        add(n+i, to, 2);
    add(to, to2, inf);
    if(m*2 > n || Dinic(from, to2) < m*2)return false;
    puts("YES");
    for(int i = 1; i <= n; i++)
    {
        ans[i] = -1;
        for(int j = head[i]; ~j; j = edge[j].nex)
        {
            if(edge[j].cap==0 && edge[j].to != from)
            {
                ans[i] = edge[j].to-n;
                D[edge[j].to-n].push_back(i);
                break;
            }
        }
        if(ans[i]==-1 && G[i]!=-1)
            D[G[i]].push_back(i);
    }
    for(int i = 1; i <= m; i++)
    {
        printf("%d", D[i].size());
        for(int j = 0; j < D[i].size(); j++)
            printf(" %d", D[i][j]);
        puts("");
    }
    return true;
}
int main(){
	while(~scanf("%d %d",&n,&m)){
        if(solve()==false)puts("NO");
        for(int i = 1; i <= m; i++)D[i].clear();
	}
	return 0;
}
/*
5 2
1 1
1 2
1 1
1 1
1 1

5 2
1 1
1 2
1 1
1 1
2 1 2



*/

SGU 242 Student's Morning 网络流（水