• 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
    ",G.dinic(1,dre(n,m)));
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/rrsb/p/7896146.html
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