POJ1466 Girls and Boys "the maximum independent set of two graphs"

Topic Links:

http://poj.org/problem?id=1466

Main topic:

there are n students, some of whom are ambiguous, only the person who knows can form a collection. Q: you can compose up to

The number of collections makes the students between these collections have nothing to do with.

Ideas:

Select M points from n plots, so that there is no edge between the M points 22, the maximum M is how much. Maximum Independence of the binary chart

Set issues. Should be the boys, girls on each side of the building two-point map to find the largest independent set, but there are only n points, no report

The number of the boys and girls. Then take n students as one side and then n students on the other. Between the people who will connect with each other

Build edges. Then the maximum number of matches is calculated. Because if you and V are connected, both sides (U,V) and (v,u) are added to the two-minute chart,

was repeated calculations for two times. And because the maximum independent set of the binary graph = N-Two The maximum match number of the graph. So the final answer is

N-Maximum number of matches/2.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace std;const int maxn = 550;bool map[maxn][maxn];bool mask[maxn];int nx,ny;int cx[maxn],cy[maxn];int FindPath (int u) {for ( int i = 0; i < NX; ++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 () {for (int i = 0; i < NX; ++i) cx[i] = 1;    for (int i = 0; i < NY; ++i) cy[i] = 1;    int res = 0;                for (int i = 0, i < NX; ++i) {if (cx[i] = = 1) {for (int j = 0; j < NY; ++j)            MASK[J] = 0;        Res + = Findpath (i); }} return res;}    int main () {int n,k,u,v;        while (~SCANF ("%d", &n)) {memset (map,0,sizeof (MAP));        NX = NY = N; for (int i = 0; i < N;            ++i) {scanf ("%d: (%d)", &u,&k);                for (int j = 0; j < K; ++j) {scanf ("%d", &v);                MAP[U][V] = 1;            Map[v][u] = 1;    }} printf ("%d\n", N-maxmatch ()/2); } return 0;}

POJ1466 Girls and Boys "the maximum independent set of two graphs"