POJ 1486 Sorting Slides 二分图关键边 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

46 22 10 204 18 6 168 20 2 1810 24 4 89 1519 1711 721 1120 2 0 20 2 0 21 11 10

Sample Output

Heap 1(A,4) (B,1) (C,2) (D,3)Heap 2none

关键边   只需要对所有的边  尝试删除这条边  删完之后做匈牙利  如果还能得到最大匹配 ans=n  那么这个边不是关键边  否则就是关键边

PS:这题我写烦了   用有向的好写多了   无向的注意数组范围  2*maxlen 别小了  

#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 30using 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]; //与i有边相连的点的集合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;}}}//forreturn 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;}}}//forreturn 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);//输入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);//构造图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); //点和面的等价化   面为0~n-1  点为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;}*///输出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);//暂时删掉//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ansG[i][j] = 1;G[j][i] = 1;//G[i].insert(it11,newp);//放回来//G[newp].insert(it22,i);if(newans != ans) //关键边{//if(ifhave == 1) printf(" ");printf("(%c,%d) ", 'A' + i, j + 1 - n);ifhave = 1;//break;}}}if(ifhave == 0)printf("none\n");elseprintf("\n");printf("\n");}//cout<<1<<endl;return 0;}