uva10092 The Problem with the Problemsetter

题意：命题人的问题，命题人被给了很多题目，每一道题目都属于某一个或几个种类，出题人要在给定的种类中选出要求数量的题目。能够选出来的打印1，并且打印每个种类选出来的书，不能选出来的打印0

分析：套用匈牙利算法，重点在于构造二部图，参照uva10080 Gopher II，将每一道题看做一个gopher,每个种类看做相同的几个洞，如果A类型需要选三道，那么就将A类型看做三个洞，编号为0 、 1 、 2.那么如果有一道题归为A类的话，在连接表或者链接矩阵中就要和0 、 1、 2都有联系。

WA经验：问题总是处在初始化link 和 visited 上，对于小数字量与会发觉，要么给到很大的范围，要么清楚地知道link 、visited 分别对于谁而言，下标范围要相应的变化。

代码：

//uva10092The Problem with the Problemsetter
//AC By Warteac
//2013-4-20
//Runtime:0.152s
#include<iostream>
#include<cmath>
#include<vector>
#include<cstring>
using namespace std;
class Problemsetter{
private:
	vector < vector<int> > selectProb;//邻接表
	int numberOfCategories;//3
	int numberOfProblems;//15
	int aimNumOfProb;//3+3+4
	int numOfMatch;
	vector <int> proNumOfEachCate;//3 3 4
	vector <int> firstIndexOfCate;
private:
	vector <int> link;
	vector <bool> visited;
    bool dfs(int u);
public:
	void initial(int,int);
	void readCase();
	void computing();
	void outResult();
};
void Problemsetter::initial(int n,int m){
	numberOfCategories = n;
	numberOfProblems = m;
	selectProb.clear();
	proNumOfEachCate.clear();
	proNumOfEachCate.push_back(0);//下标从1开始
	firstIndexOfCate.clear();
	firstIndexOfCate.push_back(0);//下标从1开始
	aimNumOfProb = 0;
	numOfMatch = 0;
	link.clear();
	visited.clear();
}
void Problemsetter::readCase(){
    int n = numberOfCategories;
	int m = numberOfProblems;
	int temp,cate;
	vector <int> category;
	while(n--){
		cin >> temp;
		aimNumOfProb += temp;
		firstIndexOfCate.push_back(firstIndexOfCate[firstIndexOfCate.size()-1]+proNumOfEachCate[proNumOfEachCate.size()-1]);
		proNumOfEachCate.push_back(temp);
		
	}
	while(m--){
		cin >> cate;
		category.clear();
		while(cate--){
			cin >> temp;
			//将每本书可以归到每一类：第一本归到第一类3次，分别编号从0到2，归到第二类3次，编号从3到5
			for(int i = 0; i < proNumOfEachCate[temp]; i++)
				category.push_back(firstIndexOfCate[temp]+i);
		}
		//将每本书的记录加入表中，每一本书是一行记录，下标从0开始
		selectProb.push_back(category);
	}
}
bool Problemsetter::dfs(int u){//visited ,link 的初始化，link的初始化只需要一次，visited每DFS一次都要初始化
	for(int i = 0; i < selectProb[u].size(); i++){
		//cout << "j = " << i <<endl;
		int c = selectProb[u][i];//cout << "c = " << c<<endl;
		if(!visited[c]){  //visited[i]-WA,dfs递归调用时就会出错了
			visited[c] = true;
			if( (link[c] == -1 ) || dfs(link[c])==true ){
				link[c] = u;
				return true;
			}
		}
	}
	return false;
}

void Problemsetter::computing(){//hungary
	for(int i = 0; i < aimNumOfProb; i++){//initial link
		link.push_back(-1);
	}
	for(int i = 0; i < selectProb.size(); i++){
		//cout << "i = " <<i<<endl;
		visited.clear();
		visited.resize(aimNumOfProb);
		if(dfs(i)) numOfMatch++;
	}
}
void Problemsetter::outResult(){
	/*for(int x = 0 ; x < selectProb.size(); x++ ){
		for(int y = 0; y < selectProb[x].size(); y++){
			cout << selectProb[x][y] << " ";
		}
		cout << endl;
	}*/
	if(numOfMatch == aimNumOfProb){
	cout << "1" <<endl;
	for(int x = 1; x <= numberOfCategories; x++){
		int t = 0;
		for (int i = 0; i < link.size(); i++){
			if( i >= firstIndexOfCate[x] && i < (firstIndexOfCate[x] + proNumOfEachCate[x])){
				if(t) cout << " "; t++;
				cout << link[i]+1;
			}
		}
		cout <<endl;
	}
	}else{ cout << "0" <<endl;}
	//cout <<"aimNumOfProb = "<<aimNumOfProb<<endl;
}
int main(){
	Problemsetter p;
	int n,m;
	while(cin >> n >> m  && n && m){
		p.initial(n,m);
		p.readCase();
		p.computing();
		p.outResult();
	}
	
}