HDU 1045 Fire Net 二分图

题意：在给出的图中‘.’放置点。（同一行或者同一列的联通块中不能放置两个或以上黑点）

思路：二分图，同一行的联通块缩成一个点，因为是等价的，称为二分图的X部，同理，同一列的联通块缩成一个点，称为二分图的Y部，‘.’处X部和Y部连一条线。找最大匹配即可。

http://acm.hdu.edu.cn/showproblem.php?pid=1045

/*********************************************    Problem : HDU 1045    Author  : NMfloat    InkTime (c) NM . All Rights Reserved .********************************************/#include <map>#include <set>#include <queue>#include <cmath>#include <ctime>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define rep(i,a,b)  for(int i = a ; i <= b ; i ++)#define rrep(i,a,b) for(int i = b ; i >= a ; i --)#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)#define cls(a,x)   memset(a,x,sizeof(a))#define eps 1e-8using namespace std;const int MOD = 1e9+7;const int INF = 0x3f3f3f3f;const int MAXN = 32;const int MAXE = 256;typedef long long LL;typedef unsigned long long ULL;struct Edge { //记录边    int to;    Edge * next;}E[MAXE],*EE;struct Gragh { //记录图的结点    Edge * first;}G[MAXN];int N,M;//二分图左右结点的个数bool visit[MAXN];int match[MAXN];//v2中匹配的情况void addedge(int u,int v) { //加边    EE->to = v ; EE -> next = G[u].first ; G[u].first = EE ++;    //EE->to = u ; EE -> next = G[v].first ; G[v].first = EE ++;}int T,n,m,k;void init() {    EE = E; N = M = 0;    cls(G,0);}bool find_path(int u) {    int v;    repE(p,u) {        v = p->to;        if(!visit[v]) {            visit[v] = 1;            if(match[v] == -1 || find_path(match[v])) {//v没有匹配或者v可以找到另一条路径                match[v] = u;                return true;            }        }    }    return false;}int Max_match() {    cls(match,-1);    int cnt = 0;    rep(i,1,N) {        cls(visit,0);        if(find_path(i)) cnt ++;    }    return cnt;}char Map[10][10];int belongv1[10][10];int belongv2[10][10];bool vis[10][10];void input() {    rep(i,1,n) scanf("%s",Map[i]+1) ;//, printf("%s\n",Map[i]+1);}void solve() {    //横排,v1    cls(vis,0); cls(belongv1,0) ; cls(belongv2,0) ;    rep(i,1,n) rep(j,1,n) {        if(!vis[i][j] && Map[i][j] == '.') {            N ++;            rep(ja,j,n) {                if(Map[i][ja] == '.') vis[i][ja] = true , belongv1[i][ja] = N;                else break;            }        }    }    //竖排,v2    cls(vis,0);    rep(i,1,n) rep(j,1,n) {        if(!vis[i][j] && Map[i][j] == '.') {            M ++;            rep(ia,i,n) {                if(Map[ia][j] == '.') vis[ia][j] = true , belongv2[ia][j] = N + M;                else break;            }        }    }    rep(i,1,n) rep(j,1,n) {        if(Map[i][j] == '.') {            addedge(belongv1[i][j],belongv2[i][j]);        }    }    //rep(i,1,n) rep(j,1,n) printf("v1 : %d v2 : %d\n",belongv1[i][j],belongv2[i][j]);    printf("%d\n",Max_match());}int main(void) {    //freopen("a.in","r",stdin);    while(scanf("%d",&n),n) {        init();        input();        solve();    }    return 0;}