[HDOJ5093] Battle ships（最大匹配）

题目链接：https://vjudge.net/problem/HDU-5093

按照行和列分别标注*的id，合并同行或同列相邻的块，二分图两部分分别是行和列，某一点(i,j)则连一条rid(i,j)到cid(i,j)的边。跑最大匹配。

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

typedef struct pair<int, int> pii;
const int maxn = 110;
const int maxm = 3030;
const int inf = 0x3f3f3f3f;

int n, m;
char mp[maxn][maxn];

int nu, nv, dist;
int Mx[maxm], My[maxm], dx[maxm], dy[maxm], vis[maxm], G[maxm][maxm];

bool Dfs(int u){
  for(int v = 1; v <= nv; v++)
    if(!vis[v] && G[u][v] && dy[v] == dx[u] + 1){
      vis[v] = 1;
      if(My[v] != -1 && dy[v] == dist)
        continue;
      if(My[v] == -1|| Dfs(My[v])){
        My[v] = u;
        Mx[u] = v;
        return 1;
      }
    }
  return 0;
}
bool Search(){
  queue<int> Q;
  dist = inf;
  memset(dx, -1, sizeof(dx));
  memset(dy, -1, sizeof(dy));
  for(int i = 1; i <= nu; i++)
    if(Mx[i] == -1){
      Q.push(i);
      dx[i] = 0;
    }
  while(!Q.empty()){
    int u = Q.front();
    Q.pop();
    if(dx[u] > dist)
      break;
    for(int v = 1; v <= nv; v++)
      if(G[u][v] && dy[v] == -1){
        dy[v] = dx[u] + 1;
        if(My[v] == -1)
          dist = dy[v];
        else{
          dx[My[v]] = dy[v] + 1;
          Q.push(My[v]); 
        }
      } 
  }
  return dist != inf;
}
int MaxMatch() {
  int res = 0;
  memset(Mx, -1, sizeof(Mx));
  memset(My, -1, sizeof(My));
  while(Search()){
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < nu; i++)
      if(Mx[i] == -1&& Dfs(i))
        res++;
  }
  return res;
}

int rid[maxn][maxn], cid[maxn][maxn];


int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d",&n,&m);
        for(int i = 0; i < n; i++) {
            scanf("%s", mp[i]);
        }
        memset(G, 0, sizeof(G));
        memset(rid, 0, sizeof(rid));
        memset(cid, 0, sizeof(cid));
        int cnt = 1;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(mp[i][j] == '*') rid[i][j] = cnt;
                if(mp[i][j] == '#') rid[i][j] = cnt++;
            }
            cnt++;
        }
        nu = cnt;
        cnt = 1;
        for(int j = 0; j < m; j++) {
            for(int i = 0; i < n; i++) {
                if(mp[i][j] == '*') cid[i][j] = cnt;
                if(mp[i][j] == '#') cid[i][j] = cnt++;
            }
            cnt++;
        }
        nv = cnt;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(mp[i][j] == '*') {
                    G[rid[i][j]][cid[i][j]] = 1;
                }
            }
        }
        printf("%d
", MaxMatch());
    }
    return 0;
}