zzulioj 2636: G（二分图）

传送门
经典二分图，经常用于解决这种覆盖问题。原理是把二维变成一维，把相邻的块当成边，然后进行二分图求最大匹配，单向边就可以。

#include<cstdio>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<math.h>
#include<climits>
#include<set>
#include<sstream>
#include<time.h>
#include<iomanip>

#define debug(x) cout <<#x<<" = "<<x<<endl
#define debug2(x, y) cout<<#x<<" = "<<x<<", "<<#y<<" = "<<y<<endl
#define gg cout <<"---------------QAQ---------------"<<endl
#define fi first
#define SZ(x) (int)x.size()
#define se second
#define pb push_back
#define MEM(a) memset(a, 0, sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define PI 3.14
#define endl "\n"
#define eps 1e-8
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<pii, ll> Pii;
template<class T> inline void read(T &x){
    x=0; char c=getchar(); int f=1;
    while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
    while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
const int N = 4e4+10, maxn = 1e6+10;
void FAST(){ios::sync_with_stdio(false);cin.tie(nullptr); cout.tie(nullptr);}
const ll mod =  1e9+7;


int n, m, k;
vector<int> e[N];
int mat[N];
int mp[210][210];
bool vis[N];

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
bool check(int x, int y)
{
    if(x<1||y<1||x>n||y>n||mp[x][y]) return false;
    return true;
}
bool dfs(int u)
{
    for(int v:e[u])
    {
        if(vis[v]) continue;
        vis[v] = 1;
        if(!mat[v]||dfs(mat[v])) {
            mat[v] = u;
            mat[u] = v;
            return true;
        }
    }
    return false;
}
void dfs1(int x, int y)
{
    for(int i = 0; i < 4; ++i) {
        int tx = x+dir[i][0], ty = y+dir[i][1];
        if(check(tx, ty)) {
            int u = x*n+y, v = tx*n+ty;
            e[u].pb(v);
        }
    }
}
void solve()
{
   scanf("%d%d", &n, &m);
   while(m--)
   {
       int x, y;
       scanf("%d%d", &x, &y);
       mp[x][y] = 1;
   }
   for(int i = 1; i <= n; ++i)
   for(int j = 1; j <= n; ++j)
   {
       if((i+j)&1||mp[i][j]) continue; 
       dfs1(i, j);
   }
   int ans = 0;
   for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= n; ++j) {
        if((i+j)&1||mp[i][j]) continue;
        memset(vis, 0, sizeof(vis));
        if(dfs(i*n+j)) ans++;
    }
    printf("%d\n", ans);
}

int main()
{
//    FAST();
//    init();
//    int _;scanf("%d", &_); while(_--)
//    while(scanf("%d", &n)&&n)
//    while(scanf("%d%d%d", &n, &m, &k)&&(n+m+k))
//    for(ll i = 1; i <= _; ++i)
    solve();
    return 0;
}