题目链接：点击打开链接
题意：
给定n个人，m个终点
以下n行表示每一个人能够去m个点。
每一个人仅仅能去一个点。
输出随意一个方案使得每一个点至少有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。推断是否满流，若不满流就无解。
若满流则其它的人随便走。
#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
*/