题解：
图就是题解。
黄色格子只能跳到红色格子上。
于是就和方格取数问题一样了。
代码：
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 250
#define ll long long
const int inf = 0x3f3f3f3f;
const ll Inf = 0x3f3f3f3f3f3f3f3fll;
inline int rd()
{
int f=1,c=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();}
return f*c;
}
int n,m,S,T,hed[N*N],cur[N*N],cnt=-1;
bool ban[N][N];
int dx[]={-2,-2,-1,-1,1,1,2,2};
int dy[]={1,-1,2,-2,2,-2,1,-1};
int _id(int x,int y)
{
return (x-1)*n+y;
}
bool check(int x,int y)
{
if(x<1||y<1||x>n||y>n)return 0;
if(ban[x][y])return 0;
return 1;
}
struct EG
{
int to,nxt;
ll w;
}e[N*N*20];
void ae(int f,int t,ll w)
{
e[++cnt].to = t;
e[cnt].nxt = hed[f];
e[cnt].w = w;
hed[f] = cnt;
}
int dep[N*N];
bool vis[N*N];
queue<int>q;
bool bfs()
{
memset(dep,0x3f,sizeof(dep));
memcpy(cur,hed,sizeof(cur));
dep[S]=0,vis[S]=1;q.push(S);
while(!q.empty())
{
int u = q.front();
q.pop();
for(int j=hed[u];~j;j=e[j].nxt)
{
int to = e[j].to;
if(e[j].w&&dep[to]>dep[u]+1)
{
dep[to] = dep[u]+1;
if(!vis[to])
{
vis[to] = 1;
q.push(to);
}
}
}
vis[u] = 0;
}
return dep[T]!=inf;
}
ll dfs(int u,ll lim)
{
if(u==T||!lim)return lim;
ll fl = 0,f;
for(int j=cur[u];~j;j=e[j].nxt)
{
cur[u] = j;
int to = e[j].to;
if(dep[to]==dep[u]+1&&(f=dfs(to,min(lim,e[j].w))))
{
fl+=f,lim-=f;
e[j].w-=f,e[j^1].w+=f;
if(!lim)break;
}
}
return fl;
}
ll dinic()
{
ll ret = 0;
while(bfs())ret+=dfs(S,Inf);
return ret;
}
int main()
{
n = rd(),m = rd();
S = n*n+1,T = n*n+2;
memset(hed,-1,sizeof(hed));
for(int x,y,i=1;i<=m;i++)
{
x = rd(),y = rd();
ban[x][y] = 1;
}
ll sum = n*n-m;
for(int i=1;i<=n;i++)
for(int u,j=1;j<=n;j++)
{
if(ban[i][j])continue;
u = _id(i,j);
if((i+j)&1)
{
ae(S,u,1);
ae(u,S,0);
for(int x,y,k=0;k<8;k++)
{
x = i+dx[k],y = j+dy[k];
if(check(x,y))
{
int to = _id(x,y);
ae(u,to,Inf);
ae(to,u,0);
}
}
}else
{
ae(u,T,1);
ae(T,u,0);
}
}
printf("%lld\n",sum-dinic());
return 0;
}