【网络流24题】骑士共存问题

题面：

https://www.luogu.org/problemnew/show/P3355

题解：

水题，染色+最小割。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<queue>
#define ri register int
#define N 50000
#define INF 1000000007
#define S 0
#define T (n*n+1)
using namespace std;

const int dx[]={-2,-2,-1,-1,1,1,2,2},dy[]={1,-1,2,-2,-2,2,1,-1};
int n,m,x,y,g[250][250];

struct graph {
  vector<int> to,w;
  vector<int> ed[N];
  int cur[N],d[N];
  void add_edge(int u,int v,int w1) {
    //cout<<u<<" "<<v<<" "<<w1<<endl;
    to.push_back(v); w.push_back(w1); ed[u].push_back(to.size()-1);
    to.push_back(u); w.push_back(0);  ed[v].push_back(to.size()-1);
  }
  bool bfs() {
    queue<int> q;
    memset(d,0x3f,sizeof(d));
    d[0]=0; q.push(0);
    while (!q.empty()) {
      int x=q.front(); q.pop();
      for (ri i=0,l=ed[x].size();i<l;i++) {
        int e=ed[x][i];
        if (w[e] && d[x]+1<d[to[e]]) {
          d[to[e]]=d[x]+1;
          q.push(to[e]);
        }
      }
    }
    return d[T]<INF;
  }
  int dfs(int x,int limit) {
    if (x==T || !limit) return limit;
    int tot=0;
    for (ri &i=cur[x];i<ed[x].size();i++) {
      int e=ed[x][i];
      if (d[to[e]]==d[x]+1 && w[e]) {
        int f=dfs(to[e],min(limit,w[e]));
        if (!f) continue;
        w[e]-=f; w[1^e]+=f; 
        tot+=f; limit-=f;
        if (!limit) return tot;
      }
    }
    return tot;
  }
  int dinic() {
    int ret=0;
    while (bfs()) {
      memset(cur,0,sizeof(cur));
      ret+=dfs(S,INF);
    }
    return ret;
  }
} G;

int main(){
  scanf("%d %d",&n,&m);
  for (ri i=1;i<=m;i++) {
    scanf("%d %d",&x,&y);
    g[x][y]=1;
  }
  int sum=0;
  for (ri i=1;i<=n;i++) 
    for (ri j=1;j<=n;j++) if (!g[i][j]) {
      sum++;
      if ((i+j)%2==0) {
        G.add_edge(S,n*(i-1)+j,1);
        for (ri k=0;k<8;k++) {
          int nx=i+dx[k],ny=j+dy[k];
          if (nx>=1 && nx<=n && ny>=1 && ny<=n && g[nx][ny]==0) G.add_edge(n*(i-1)+j,n*(nx-1)+ny,INF);
        }
      }
      else G.add_edge(n*(i-1)+j,T,1);
    }
  cout<<sum-G.dinic()<<endl;
}