BZOJ 1001 sol

本题让我们求这张图的最小割，最小割等于最大流

下副代码

#include<bits/stdc++.h>
#define eho(x) for(int& i=hed[x];~i;i=net[i])
#define Eho(x) for(int i=head[x];~i;i=net[i])
#define N 1000007
#define M 6000007
#define dre(x,y) (((x)-1)*(m)+(y))
#define sight(c) (c<='9'&&c>='0')
#define INF (1<<20)
using namespace std;
int n,m,C;
char c;
void read(int &x){
    c=getchar();
    for(;!sight(c);c=getchar());
    for(x=0;sight(c);c=getchar()) x=(x<<3)+(x<<1)+c-48;
}
struct G{
    int head[N],net[M],fall[M],cost[M],s,t,tot,d[N],hed[N];
    bool in[N];
    queue<int> Q;
    G() { memset(head,-1,sizeof head);tot=-1;}
    inline void add(int x,int y,int c){
        fall[++tot]=y; net[tot]=head[x]; head[x]=tot; cost[tot]=c;
    }
    inline void adds(int x,int y,int c){
        add(x,y,c); add(y,x,c);}
    inline bool spfa(){
        memset(d,127,sizeof d); int x;
        d[s]=1; Q.push(s); in[s]=1;
        while (!Q.empty()) {
            x=Q.front(); Q.pop();
            Eho(x)
                if (cost[i]&&d[fall[i]]>d[x]+1) {
                    d[fall[i]]=d[x]+1;
                    if (!in[fall[i]]) {in[fall[i]]=1,Q.push(fall[i]);}
                }
            in[x]=0;
        }
       return d[t]<INF;
    }
    inline int dfs(int x,int F){
        if (x==t|| !F) return F;
        int flow=0,r;
        eho(x)
         if (d[x]+1==d[fall[i]]&&((r=dfs(fall[i],min(F,cost[i])))>0)) {
          cost[i]-=r;
          cost[i^1]+=r;
          F-=r; flow+=r; if (!F) break;}
        return flow;    
    }
    int dinic(int A,int B){
        s=A; t=B;
        int flow=0;
        while (spfa()) {
         memcpy(hed,head,sizeof head);
         flow+=dfs(s,INF); }
        return flow;
    }
}G;
int main () {
    read(n); read(m);
    for (int i=1;i<=n;i++)
     for (int j=1;j<m;j++) 
         read(C),G.adds(dre(i,j),dre(i,j+1),C);
    for (int i=1;i<n;i++)
     for (int j=1;j<=m;j++)
        read(C),G.adds(dre(i,j),dre(i+1,j),C);
    for (int i=1;i<n;i++)
     for (int j=1;j<m;j++)
        read(C),G.adds(dre(i,j),dre(i+1,j+1),C);
    printf("%d\n",G.dinic(1,dre(n,m)));
    return 0;
}