King's Quest
Once upon a time there lived a king and he had N sons. And there were N beautiful girls in the kingdom and the king knew about each of his sons which of those girls he did like. The sons of the king were young and light-headed, so it was possible for one son to like several girls.
So the king asked his wizard to find for each of his sons the girl he liked, so that he could marry her. And the king's wizard did it -- for each son the girl that he could marry was chosen, so that he liked this girl and, of course, each beautiful girl had to marry only one of the king's sons.
However, the king looked at the list and said: "I like the list you have made, but I am not completely satisfied. For each son I would like to know all the girls that he can marry. Of course, after he marries any of those girls, for each other son you must still be able to choose the girl he likes to marry."
The problem the king wanted the wizard to solve had become too hard for him. You must save wizard's head by solving this problem.
Input
The first line of the input contains N -- the number of king's sons (1 <= N <= 2000). Next N lines for each of king's sons contain the list of the girls he likes: first Ki -- the number of those girls, and then Ki different integer numbers, ranging from 1 to N denoting the girls. The sum of all Ki does not exceed 200000.
The last line of the case contains the original list the wizard had made -- N different integer numbers: for each son the number of the girl he would marry in compliance with this list. It is guaranteed that the list is correct, that is, each son likes the girl he must marry according to this list.
Output
Output N lines.For each king's son first print Li -- the number of different girls he likes and can marry so that after his marriage it is possible to marry each of the other king's sons. After that print Li different integer numbers denoting those girls, in ascending order.
Sample Input
4
2 1 2
2 1 2
2 2 3
2 3 4
1 2 3 4
Sample Output
2 1 2
2 1 2
1 3
1 4
Hint
This problem has huge input and output data,use scanf() and printf() instead of cin and cout to read data to avoid time limit exceed.
Source
题目类型：有向图强连通分量(二分图完备匹配)
算法分析：这道题直观上是一道求解二分图完备匹配并输出所有满足条件的问题。但是由于顶点数和边数比较多，所以直接使用找增广链的算法会超时。这里利用二分图完备匹配的性质来简化问题。如何才能不通过寻找增广链而直接确定是否存在增广链呢？让我们再次回到初始的想法，如果给定的完备匹配中xi -> yi，现在我们想让xi指向yj。如果让xi -> yj的话，那么yi显然对于xi已经没用了，又因为最终的匹配必然是一个完备匹配，即yi一定会被某一个xk(1 <= k <= n)所匹配。所以假如寻找到了增广链，那么增广链的最后一条边所指向的顶点必然是yi。假如我们由yi向xi再连一条边会发生什么呢？ 很明显，形成了一个环！也就是说，形成了一个强连通分量。所以说，如果xi可以选择yj(即xi喜欢yi且可以结婚)，那么xi和yj必然属于同一个强连通分量。所以此题可用强连通分量来求解，求解出强连通分量后直接统计即可
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/************************************************* Author :supermaker Created Time :2016/1/25 12:38:12 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 2 * 2000 + 66; int dfn[maxn], low[maxn], belong[maxn]; int n, id, tot; bool instack[maxn], gg[maxn][maxn]; vector <int> g[maxn]; stack <int> sta; void dfs (int u) { dfn[u] = low[u] = ++id; sta.push (u); instack[u] = true; for (int i = 0; i < g[u].size (); i++) { int v = g[u][i]; if (!dfn[v]) { dfs (v); low[u] = min (low[u], low[v]); } else if (instack[v]) low[u] = min (low[u], dfn[v]); } if (low[u] == dfn[u]) { tot++; int tt; do { tt = sta.top (); sta.pop (); instack[tt] = false; belong[tt] = tot; } while (!sta.empty () && tt != u); } } void SS () { memset (dfn, 0, sizeof (dfn)); memset (low, 0, sizeof (low)); memset (instack, false, sizeof (instack)); id = tot = 0; while (!sta.empty ()) sta.pop (); for (int i = 1; i <= n; i++) if (!dfn[i]) dfs (i); } int main() { //CFF; //CPPFF; while (scanf ("%d", &n) != EOF) { for (int i = 0; i < maxn; i++) g[i].clear (); memset (gg, false, sizeof (gg)); for (int i = 1; i <= n; i++) { int num; scanf ("%d", &num); for (int j = 1; j <= num; j++) { int u; scanf ("%d", &u); g[i].push_back (n + u); gg[i][u] = true; } } for (int i = 1; i <= n; i++) { int u; scanf ("%d", &u); g[n+u].push_back (i); } SS (); vector <int> temp; for (int i = 1; i <= n; i++) { temp.clear (); for (int j = n + 1; j <= 2 * n; j++) if (gg[i][j-n] && belong[i] == belong[j]) temp.push_back (j - n); sort (temp.begin (), temp.end ()); printf ("%d", temp.size ()); for (int j = 0; j < temp.size (); j++) printf (" %d", temp[j]); puts (""); } } return 0; } |
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