网络流二十四题之骑士共存问题

终于把这个系列刷完了，虽然还有几道题没有搞定，但是在建模上已经有了长足的进步。
这道题的建模和之前的方格取数有点像，同样先将棋盘染色。
1.源点向白色块连接一条边，容量为1，黑色块向汇点连接一条边，容量为1.
2.对于所有的白块，向它能走到的地方连一条边，容量为1
然后求一个最大流，又可以走的点减去最大流的值，其实要求最小割。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN = 200010;//点数的最大值
const int MAXM = 500010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
    int from,to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{
    edge[tol].to = v;
    edge[tol].from = u;
    edge[tol].cap = w;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].from = v;
    edge[tol].cap = rw;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while(front != rear)
    {
        int u = Q[front++];
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(dep[v] != -1)
                continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}
int S[MAXN];
int sap(int start,int end,int N)
{
    BFS(start,end);
    memcpy(cur,head,sizeof(head));
    int top = 0;
    int u = start;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            int inser;
            for(int i = 0; i < top; i++)
                if(Min > edge[S[i]].cap - edge[S[i]].flow)
                {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            for(int i = 0; i < top; i++)
            {
                edge[S[i]].flow += Min;
                edge[S[i]^1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top]^1].to;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = true;
                cur[u] = i;
                break;
            }
        }
        if(flag)
        {
            S[top++] = cur[u];
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])
            return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if(u != start)
            u = edge[S[--top]^1].to;
    }
    return ans;
}
void print_all()
{
    for(int i = 0; i<tol; i+=2)
    {
        printf("%d->%d cap = %d flow = %d\n",edge[i].from,edge[i].to,edge[i].cap,edge[i].flow);
    }
}
int dx[] = {-2,-2,-1,-1,1,1,2,2};
int dy[] = {1,-1,2,-2,-2,2,1,-1};
int id[201][201];
int c[201][201];
int block[201][201];
int main()
{

    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 1; i<=m; i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        block[x][y] = 1;
    }
    init();
    int b = 1,w = (n *n +1)/2+1;
    for(int i = 1;i<=n;i++)
    {
        for(int j = 1;j<=n;j++)
        {
            if( (i + j) % 2)
            {
                id[i][j] = b++;
            }
            else
            {
                id[i][j] = w++;
            }
        }
    }
    for(int i = 1;i<=n;i++)
    {
        for(int j  =1;j<=n;j++)
        {
            if(block[i][j]) continue;
            else if(id[i][j] < b)
            {
                for(int k = 0;k<8;k++)
                {
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if(x <1 || y< 1|| x > n || y> n || block[x][y]) continue;
                    addedge(id[i][j],id[x][y],1);
                }
                addedge(0,id[i][j],1);
            }
            else
            {
                addedge(id[i][j],w,1);
            }
        }
    }
    int ans = sap(0,w,w+1);
    printf("%d\n",n*n - m-ans);
    return 0;
}