Gym 101201G Maximum Islands (最大独立集)

题意：给定一个图，L代表陆地，W代表水，C表示不确定，问你最多有多少岛。

析：首先给定的L周围必须是是W，只有这样才是最优的，因为如果是L，那么还得有另外的W来包围，不是最优的，那么剩下的就剩下C了，因为要是L多，那么肯定是一个岛屿只有一个L，这样是最优的，并且周围都是W，所以可以把C看成一个点，然后向周围上下左右连边，如果周围存在C，那么就连一条，最后求一个最大独立集就OK了，二分图的最大独立集等于二分图的顶点数 - 二分图的最大匹配。也就是求二分匹配。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 50 + 50;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
  int to, next;
};
Edge edge[maxn*maxn];
int head[maxn*maxn], cnt;
int G[maxn][maxn];
char s[maxn][maxn];

void addEdge(int u, int v){
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
bool vis[maxn][maxn];
bool used[maxn*maxn];
int match[maxn*maxn];

void dfs(int r, int c){
  for(int i = 0; i < 4; ++i){
    int x = r + dr[i];
    int y = c + dc[i];
    if(vis[x][y] || !is_in(x, y))  continue;
    vis[x][y] = 1;
    if(s[x][y] == 'L')  dfs(x, y);
    else s[x][y] = 'W';
  }
}

bool dfs(int u){
  used[u] = true;
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to, w = match[v];
    if(w < 0 || !used[w] && dfs(w)){
      match[u] = v;
      match[v] = u;
      return true;
    }
  }
  return false;
}

int main(){
  scanf("%d %d", &n, &m);
  for(int i = 0; i < n; ++i)
    scanf("%s", s[i]);
  int ans = 0;
  FOR(i, 0, n)  FOR(j, 0, m)
    if(!vis[i][j] && s[i][j] == 'L'){
      vis[i][j] = 1;
      dfs(i, j);  ++ans;
    }
  ms(G, -1);  ms(head, -1);  cnt = 0;
  int idx = 0;
  FOR(i, 0, n)  FOR(j, 0, m)
    if(s[i][j] == 'C')  G[i][j] = idx++;
  FOR(i, 0, n)  FOR(j, 0, m)
    if(i+j&1&&~G[i][j]){
      for(int k = 0; k < 4; ++k){
        int x = dr[k] + i;
        int y = dc[k] + j;
        if(is_in(x, y) && ~G[x][y]){
          addEdge(G[i][j], G[x][y]);
          addEdge(G[x][y], G[i][j]);
        }
      }
    }
  ms(match, -1);
  int cnt = 0;
  for(int i = 0; i < idx; ++i)  if(match[i] < 0){
    ms(used, 0);  if(dfs(i))  ++cnt;
  }
  printf("%d\n", ans += idx - cnt);
  return 0;
}