POJ 1486 Sorting Slides 二分圖關鍵邊 匈牙利算法

Professor Clumsey is going to give an important talk this afternoon. Unfortunately, he is not a very tidy person and has put all his transparencies on one big heap. Before giving the talk, he has to sort the slides. Being a kind of minimalist, he wants to do this with the minimum amount of work possible. 

The situation is like this. The slides all have numbers written on them according to their order in the talk. Since the slides lie on each other and are transparent, one cannot see on which slide each number is written. 

Well, one cannot see on which slide a number is written, but one may deduce which numbers are written on which slides. If we label the slides which characters A, B, C, ... as in the figure above, it is obvious that D has number 3, B has number 1, C number 2 and A number 4. 

Your task, should you choose to accept it, is to write a program that automates this process.

Input

The input consists of several heap descriptions. Each heap descriptions starts with a line containing a single integer n, the number of slides in the heap. The following n lines contain four integers xmin, xmax, ymin and ymax, each, the bounding coordinates of the slides. The slides will be labeled as A, B, C, ... in the order of the input. 

This is followed by n lines containing two integers each, the x- and y-coordinates of the n numbers printed on the slides. The first coordinate pair will be for number 1, the next pair for 2, etc. No number will lie on a slide boundary. 

The input is terminated by a heap description starting with n = 0, which should not be processed. 

Output

For each heap description in the input first output its number. Then print a series of all the slides whose numbers can be uniquely determined from the input. Order the pairs by their letter identifier. 

If no matchings can be determined from the input, just print the word none on a line by itself. 

Output a blank line after each test case. 

Sample Input

4
6 22 10 20
4 18 6 16
8 20 2 18
10 24 4 8
9 15
19 17
11 7
21 11
2
0 2 0 2
0 2 0 2
1 1
1 1
0

Sample Output

Heap 1
(A,4) (B,1) (C,2) (D,3)

Heap 2
none

關鍵邊   只需要對所有的邊  嘗試刪除這條邊  刪完之後做匈牙利  如果還能得到最大匹配 ans=n  那麼這個邊不是關鍵邊  否則就是關鍵邊

PS:這題我寫煩了   用有向的好寫多了   無向的注意數組範圍  2*maxlen 別小了  

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<map>
#include<string>
#include<cmath>
#define ff(i,x,y) for(int i=x;i<y;i++)
#define maxlen 30

using namespace std;

int n;

struct nodepoint
{
	int x;
	int y;
};

struct nodess
{
	struct nodepoint lu;
	struct nodepoint rd;
};

nodess ss[maxlen];
nodepoint points[maxlen];
//vector<int> G[2 * maxlen]; //與i有邊相連的點的集合
int G[2 * maxlen][2 * maxlen];
int check[2*maxlen];
int matching[2*maxlen];

/*
bool dfs(int x)
{
	for(int i = 0; i < G[x].size(); i++)
	{
		int newpos = G[x][i];
		if(check[newpos] == 0)
		{
			check[newpos] = 1;
			if(matching[newpos] == -1 || dfs(matching[newpos]) == 1)
			{
				matching[newpos] = x;
				matching[x] = newpos;
				return true;
			}
		}
	}//for
	return false;
}*/

bool dfs(int x)
{
	for(int i = 0; i < 2 * n; i++)
	{
		if(G[x][i] == 1 && check[i] == 0)
		{
			check[i] = 1;
			if(matching[i] == -1 || dfs(matching[i])==1)
			{
				matching[i] = x;
				matching[x] = i;
				return true;
			}
		}
	}//for
	return false;
}


int main()
{
	//freopen("test.txt", "r", stdin);
	int key = 0;
	while(1)
	{
		key++;
		memset(G, 0, sizeof G);
		//for(int i = 0; i < 2 * maxlen; i++)
		//G[i].resize(0);
		//輸入
		scanf("%d", &n);
		if(n == 0)
			break;

		for(int i = 0; i < n; i++)
		{
			scanf("%d %d %d %d", &ss[i].lu.x, &ss[i].rd.x, &ss[i].lu.y, &ss[i].rd.y);
		}
		for(int i = 0; i < n; i++)
			scanf("%d %d", &points[i].x, &points[i].y);


		//構造圖
		for(int i = 0; i < n; i++)
		{
			for(int j = 0; j < n; j++)
			{
				if((points[j].x >= ss[i].lu.x && points[j].x <= ss[i].rd.x) && (points[j].y >= ss[i].lu.y && points[j].y <= ss[i].rd.y))
				{
					//G[i].push_back(j + n); //點和麪的等價化   面爲0~n-1  點爲n~2n-1
					//G[j + n].push_back(i);
					G[i][j + n] = 1;
					G[j + n][i] = 1;
				}
			}
		}

		memset(matching, -1, sizeof matching);
		int ans=0;
		for(int k = 0; k < 2 * n; k++)
		{
			if(matching[k] == -1)
			{
				memset(check, 0, sizeof check);
				check[k] = 1;
				if(dfs(k) == true)
					ans++;
			}
		}//求新的ans

		//cout<<n<<" "<<ans<<endl;

		/*
		for(int i = 0; i < n; i++)
		{
			for(int j = n; j < 2*n; j++)
			{
				cout<<G[i][j]<<" ";
			}
			cout<<endl;
		}*/

		//輸出
		printf("Heap %d\n", key);

		bool ifhave = 0;
		for(int i = 0; i < n; i++)
		{
			for(int j = n; j < 2 * n; j++)
			{
				if(G[i][j] == 0)
					continue;

				//int newp = G[i][j];

				//vector<int>::iterator it11 = find(G[i].begin(), G[i].end(), newp);
				//vector<int>::iterator it22 = find(G[newp].begin(), G[newp].end(), i);
				G[i][j] = 0;
				G[j][i] = 0;

				//G[i].erase(it11);//暫時刪掉
				//G[newp].erase(it22);

				memset(matching, -1, sizeof matching);
				int newans = 0;
				for(int k = 0; k < 2 * n; k++)
				{
					if(matching[k] == -1)
					{
						memset(check, 0, sizeof check);
						//check[k] = 1;
						if(dfs(k) == 1)
							newans++;
					}
				}//求新的ans
				G[i][j] = 1;
				G[j][i] = 1;

				//G[i].insert(it11,newp);//放回來
				//G[newp].insert(it22,i);

				if(newans != ans) //關鍵邊
				{
					//if(ifhave == 1) printf(" ");
					printf("(%c,%d) ", 'A' + i, j + 1 - n);
					ifhave = 1;
					//break;
				}
			}
		}

		if(ifhave == 0)
			printf("none\n");
		else
			printf("\n");
		printf("\n");

	}
	//cout<<1<<endl;

	return 0;
}