## Solución al problema número ANGELS de Spoj - ANGELS.

Cualquier duda no dudes en contactar.

``````#include <bits/stdc++.h>
using namespace std;

const int SINK = 50001;
const int MAX_N = 50005;
const int INF = 0x3f3f3f3f;

int N, M;
int pair_g1[MAX_N], pair_g2[MAX_N], dist[MAX_N];

bool bfs() {
queue< int > q; // queue of G1 nodes

for (int v = 0; v < N; ++v) {
if (pair_g1[v] == SINK) {
dist[v] = 0;
q.push(v);
}
else {
dist[v] = INF;
}
}
dist[SINK] = INF;

while (!q.empty()) {
int v = q.front();
q.pop();

if (dist[v] < dist[SINK]) {
for (list< int >::iterator it = adj[v].begin(); it != adj[v].end(); ++it) {
int u = *it;

if (dist[pair_g2[u]] == INF) {
dist[pair_g2[u]] = dist[v] + 1;
q.push(pair_g2[u]);
}
}
}
}
return dist[SINK] != INF;
}

bool dfs(int u) {
// DFS starting at a node u in G1

if (u == SINK) {
// we've reached a free node in G2 (i.e. one pointing to SINK)
return true;
}

for (list< int >::iterator it = adj[u].begin(); it != adj[u].end(); ++it) {
int v = *it;

if (dist[pair_g2[v]] == dist[u] + 1) {
if (dfs(pair_g2[v])) {
pair_g1[u] = v;
pair_g2[v] = u;
return true;
}
}
}
dist[u] = INF;
return false;
}
//Bipartite Maching Problem -> O(sqrt(V)*e)
int hopcroft_karp() {
int matching = 0;

for (int i = 0; i < N; ++i) {
pair_g1[i] = SINK;
}
for (int j = 0; j < M; ++j) {
pair_g2[j] = SINK;
}

while (bfs()) {
for (int i = 0; i < N; ++i) {
if (pair_g1[i] == SINK) {
matching += dfs(i);
}
}
}
return matching;
}
int main() {
int n,m,cases; scanf("%d",&cases);
while(cases--)
{
scanf("%d%d",&n,&m);
for (int i = 0; i < N; i++) adj[i].clear();
cin.ignore();
char arr[n][m];
int hori[n][m],vert[n][m];
string car;
for(int i=0; i<n;i++)
for(int j=0;j<m;j++) { cin>>car; arr[i][j]=car[0];}
for(int i=0; i<n;i++)
{
for(int j=0;j<m;j++)
{
if(arr[i][j]=='A')
{
{
}
}
else if(arr[i][j]=='D') continue;
}
}
for(int i=0; i<m;i++)
{
for(int j=0;j<n;j++)
{
if(arr[j][i]=='A')
{
{
}
}
else if(arr[j][i]=='D') continue;
}
}
N=countver; M=counthor;
for(int i=0; i<n;i++)
{
for(int j=0;j<m;j++)
{
if(arr[i][j]=='H')
{
int val = vert[i][j];