【codevs1922】[網絡流24題]騎士共存問題

很輕鬆可以看出是割，然後由於馬走路的性質，可以黑白染色轉化成二分圖，跟方格取數一樣做
由源點向黑點引邊
由黑點向能跳到的點引邊
由白點向匯點引邊
直接求割

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>  
#include<queue>  
using namespace std;  
int n,m,a[203][203],point[800007],tot=-1;  
int num[800007],deep[800007],cur[800007],laste[800007];  
int next[2000007],v[2000007],remain[2000007];  
int x[10]={1,1,2,2,-1,-1,-2,-2},y[10]={2,-2,1,-1,2,-2,1,-1};  
const int inf=1e9;  
void add(int x,int y,int z)  
{  
  tot++; next[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=z;  
  tot++; next[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;  
}  
int addflow(int s,int t)  
{  
  int minn=inf;int now=t;  
  while(now!=s)  
   {  
    minn=min(minn,remain[laste[now]]);  
    now=v[laste[now]^1];  
   }  
  now=t;  
  while(now!=s)  
   {  
    remain[laste[now]]-=minn;  
    remain[laste[now]^1]+=minn;  
    now=v[laste[now]^1];  
   }  
  return minn;  
}  
void bfs(int s,int t)  
{  
  for (int i=s;i<=t;i++)  deep[i]=t;  
  deep[t]=0; queue<int> p; p.push(t);  
  while(!p.empty())  
   {  
    int now=p.front(); p.pop();  
    for (int i=point[now];i!=-1;i=next[i])  
    if (deep[v[i]]==t&&remain[i^1])  
     deep[v[i]]=deep[now]+1,p.push(v[i]);  
   }  
}  
int  isap(int s,int t)  
{  
int ans=0,now=s;  
bfs(s,t);  
for (int i=s;i<=t;i++) num[deep[i]]++;  
for (int i=s;i<=t;i++) cur[i]=point[i];  
while(deep[s]<t)  
{  
if (now==t)  
{  
ans+=addflow(s,t);  
now=s;  
}  
bool p=false;  
for (int i=cur[now];i!=-1;i=next[i])  
{  
if (deep[v[i]]+1==deep[now]&&remain[i])  
{  
laste[v[i]]=i;  
cur[now]=i;  
p=true;  
now=v[i];  
break;  
}  
}  
if (!p)  
{  
int minn=t-1;  
for (int i=point[now];i!=-1;i=next[i])  
if (remain[i]) minn=min(minn,deep[v[i]]);  
if (!--num[deep[now]]) break;  
num[deep[now]=minn+1]++;  
cur[now]=point[now];  
if (now!=s) now=v[laste[now]^1];  
}  
}  
return ans;  
}  
int main()  
{  
  memset(next,-1,sizeof(next));  
  memset(point,-1,sizeof(point));  
  scanf("%d%d",&n,&m);  
  for (int i=1;i<=m;i++)  
   {  
    int x,y; scanf("%d%d",&x,&y);  
    a[x][y]=1;  
   }  
  int num=n*n+1;  
  for (int i=1;i<=n;i++)  
   if (i%2==1)  
    {  
    for (int j=1;j<=n;j++)  
    if (!a[i][j])  
    if (j%2==1)  
     {  
      int t=(i-1)*n+j;  
      add(0,t,1);  
       for (int k=0;k<8;k++)  
{  
int xx=i+x[k]; int yy=j+y[k];  
if (xx>0&&xx<=n&&yy>0&&yy<=n&&!a[xx][yy])  
add(t,(i+x[k]-1)*n+j+y[k],inf);  
}   
     }  
    else  add((i-1)*n+j,num,1);  
    }  
   else  
    {  
      for (int j=1;j<=n;j++)  
    if (!a[i][j])  
    if (j%2==0)  
     {  
      int t=(i-1)*n+j;  
      add(0,t,1);  
       for (int k=0;k<8;k++)  
{  
int xx=i+x[k]; int yy=j+y[k];  
if (xx>0&&xx<=n&&yy>0&&yy<=n&&!a[xx][yy])  
add(t,(i+x[k]-1)*n+j+y[k],inf);  
}   
     }  
    else  add((i-1)*n+j,num,1);  

    }  
  printf("%d",n*n-isap(0,num)-m);  
}