HPU1179 Ollivanders:makers of Fine wands since 382 BC. "Binary graph Max Match"

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=1179

Main topic:

The title is too long, the simple meaning is: There are n wands, M wizards, wands have a number of matching wizards. But a magician.

Only one wand can be matched. So the question is: how many magicians can get the magic wand at most.

Ideas:

Make a dichotomy, one side is a wand, the other is a magician. The corresponding match as the side of the binary graph. Using the Hungarian algorithm, find out two

What is the maximum matching of the graph?

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace std;const int maxn = 110;bool map[maxn][maxn],mask[maxn];int n,m,cx[maxn],cy[maxn];int FindPath (int u) {for (int i = 1; I <= M;            ++i) {if (Map[u][i] &&!mask[i]) {mask[i] = 1; if (cy[i] = =-1 | |                Findpath (Cy[i])) {Cy[i] = u;                Cx[u] = i;            return 1; }}} return 0;}    int Maxmatch () {int ret = 0;    for (int i = 1; I <= N; ++i) cx[i] = 1;    for (int i = 1; I <= M; ++i) cy[i] = 1;                for (int i = 1; I <= N, ++i) {if (cx[i] = = 1) {for (int j = 1; j <= M; ++j)            MASK[J] = 0;        RET + = Findpath (i); }} return ret;}    int main () {int d,w;        while (~SCANF ("%d%d", &n,&m)) {memset (map,false,sizeof (MAP));      for (int i = 1; I <= M; ++i) {      scanf ("%d", &d);                while (d--) {scanf ("%d", &w);            Map[w][i] = 1;    }} printf ("%d\n", Maxmatch ()); } return 0;}

HPU1179 Ollivanders:makers of Fine wands since 382 BC. "Binary graph Max Match"