該題求的是 最大獨立集團權值和 因為所取的點沒有兩點是相鄰的

所以把相鄰點連起來！

由前面二分匹配知 最大獨立集團=頂點數-最小覆蓋點 加上權值也是一樣的

所以最大獨立集團權值和 ==總權值-最小覆蓋點權值和

所以求出最小覆蓋點權值和即可 且=最小割=最大流

奇偶建圖

源點-奇數 奇數和所有相鄰的建圖 偶數 -匯點

最大流=最小割=最小點權覆蓋集=sum-最大點權獨立集！

#include <iostream>
#include <queue>
#include <cstdio>
#include <algorithm>
#include
 
<cstring>
using namespace std;
const int maxn = 50+10;
const int E_maxn = 50000 + 10;
const int INF = 0x3fffffff;
struct Edge
{
    int from;
    int to;
    int cap;
    int next;
};
struct Edge edge[E_maxn];
int deep[maxn*maxn];
int cur[maxn*maxn];
int k,n,m;
int temp[4][2] = {-1,0, 1,0, 0,-1, 0,1};
void
 
 addedge(int from,int to,int cap)
{
    edge[k].from = from;
    edge[k].to = to;
    edge[k].cap = cap;
    edge[k].next = cur[from];
    cur[from] = k;
    k++;
    edge[k].from = to;
    edge[k].to = from;
    edge[k].cap = 0;
    edge[k].next = cur[to];
    cur[to] = k;
    k++;
}
int judge(int x,int
 
 y)
{
    if(x>0 && x<=n && y>0 && y<=m)
    {
        return 1;
    }
    return 0;
}
bool BFS()
{
    memset(deep,-1,sizeof(deep));
    queue<int> Q;
    Q.push(0);
    deep[0] = 0;
    while(!Q.empty())
    {
        int x = Q.front();
        Q.pop();
        int i;
        for(i=cur[x]; i!=-1; i=edge[i].next)
        {
            int e = edge[i].to;
            if(deep[e] == -1 && edge[i].cap > 0)
            {
                deep[e] = deep[x] + 1;
                Q.push(e);
            }
        }
    }
    return deep[n*m+1] != -1;
}
int DFS(int x,int a)
{
    if(x == n*m+1)
    {
        return a;
    }
    int flow = 0;
    for(int i=cur[x]; i!=-1 && flow<a; i=edge[i].next)
    {
        int e = edge[i].to;
        if(deep[x] + 1 == deep[e] && edge[i].cap > 0)
        {
            int f = min(edge[i].cap,a-flow);
            f = DFS(e,f);
            flow += f;
            edge[i].cap -= f;
            edge[i^1].cap += f;
        }
    }
    if(!flow)
    {
        deep[x] = -2;//沒有必要再往這個點走了
    }
    return flow;
}
int Dinic()
{
    int flow = 0;
    int F=0;
    while(BFS())
    {
        while(F=DFS(0,INF))
        {
            flow += F;
        }
    }
    return flow;
}
int main()
{
    while(cin>>n)
    {
        int i;
        int j;

        memset(cur,-1,sizeof(cur));
        k=0;
        int sum = 0;
        int num;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                scanf("%d",&num);
                sum+=num;
                if((i+j) % 2 == 0)
                {
                    addedge(0,(i-1)*n+j,num);
                    for(int z = 0; z<4; z++)
                    {
                        int x = i+temp[z][0];
                        int y = j+temp[z][1];
                        if(judge(x,y))
                        {
                            addedge((i-1)*n+j,(x-1)*n+y,INF);
                        }
                    }
                }
                else
                {
                    addedge((i-1)*n+j,n*n+1,num);

                }
            }
        }
        int ans = Dinic();
        cout<<sum-ans<<endl;
    }
    return 0;
}

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