洛谷 P2774 方格取数问题 二分图最大点独立集

                      洛谷  P2774 方格取数问题  最小割 -最大流

                                                                     题解1 

                                                                     题解2

  先染色成二分图

  最大取和 = 最大和-最小舍弃和 = 最小割                 二分图 ： 最大点独立集 = 总权和 -  最小点覆盖集（最小割）

#include <bits/stdc++.h>

using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 20000 + 10;
const int maxm = 1000000 + 10;

int n,m,k;
int l[maxn];//记录层数
int h[maxn];//链式前向星
int cur[maxn];
int tot = 0;

struct edge
{
  int to;
  int c;
  int next;
  edge(int x = 0, int y = 0, int z = 0) : to(x), c(y), next(z) {}
 }es[maxm*2];//记录边 注意是2倍

void add_edge(int u, int v, int c)
{
    es[tot] = edge(v,c,h[u]);
    h[u] = tot++;
    es[tot] = edge(u,0,h[v]);
    h[v] = tot++;
   // cout << u <<" " <<v << " " << c<<endl;
}

bool bfs(int s, int t)
{
   memset(l,0,sizeof(l));
   l[s] = 1;
   queue <int> q;
   q.push(s);
   while(!q.empty())
   {
    int u = q.front();
    //cout  << u <<endl;
    q.pop();
    if(u == t)  return true;
    for(int i = h[u]; i != -1; i = es[i].next)
        {
         int v = es[i].to;
         if(!l[v] && es[i].c) {l[v] = l[u] + 1; q.push(v);}
        }
   }
   return false;
}

int dfs(int x, int t, int mf)
{
    if(x == t) return mf;
    int ret = 0;
    for(int &i = cur[x]; i != -1; i = es[i].next)
    {
      if(es[i].c && l[x] == l[es[i].to] - 1)
      {
        int f = dfs(es[i].to,t,min(es[i].c,mf - ret));
        es[i].c -= f;
        es[i^1].c += f;
        ret += f;
        if(ret == mf) return ret;
      }
    }
    return ret;
}

int dinic(int s, int t)
{
  int ans = 0;
  while(bfs(s,t))
  {
   for(int i = 0; i <= t; i++) cur[i] = h[i];
   ans += dfs(s,t,INF);
   }
  return ans;
}


int r,c;

int a[maxn][maxn];

bool check(int x, int y)
{
    if(x < 1 || y < 1 || x > r || y > c) return false;
    return true;
}
int fx[4] = {1,-1,0,0};
int fy[4] = {0,0,1,-1};

int main()
{
    int ans = 0;
    tot = 0;
    memset(h,-1,sizeof(h));
    scanf("%d%d",&r,&c);
    for(int i = 1; i <= r; i++)
        for(int j = 1; j <= c; j++)
          {scanf("%d",&a[i][j]);ans += a[i][j];}
    int s = 0,t = r*c+1;
    for(int i = 1; i <= r; i++)
        for(int j = 1; j <= c; j++)
          if((i+j)%2==1)
              add_edge((i-1)*c+j,t,a[i][j]);
          else {
            add_edge(s,(i-1)*c+j,a[i][j]);
            for(int k = 0; k < 4; k++)
            {
                int fi = i+fx[k];
                int fj = j+fy[k];
                if(check(fi,fj))  add_edge((i-1)*c+j,(fi-1)*c+fj,INF);
            }
          }

     int res = dinic(s,t);
     ans -= res;
     printf("%d\n",ans);
     return 0;
}