HDU 3549 Flow Problem (网络流板子)

Dinic 邻接表

(抄袭神牛的板子 传送门:https://blog.csdn.net/u013480600/article/details/38796521)

代码:

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <math.h>
#include <vector>
#include <string>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define zero(a) fabs(a)<eps
#define lowbit(x) (x&(-x))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define MOD 1000000007
int max(int x,int y){return x>y?x:y;};
int min(int x,int y){return x<y?x:y;};
int max(int x,int y,int z){return max(x,max(y,z));};
int min(int x,int y,int z){return min(x,min(y,z));};
typedef long long LL;
const double PI=acos(-1.0);
const double eps=1e-8;
const int inf=0x3f3f3f3f;
const LL linf=0x3f3f3f3f3f3f3f3fLL;
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

using namespace std;
const int maxn=15+5;
const int maxm=2000+10;

struct Edge
{
    Edge(){}
    Edge(int from,int to,int cap,int flow):from(from),to(to),cap(cap),flow(flow){}
    int from,to,cap,flow;
};

struct Dinic
{
    int n,m,s,t;                //结点数,边数(包括反向弧),源点与汇点编号
    Edge edges[maxm];           //边表 edges[e]和edges[e^1]互为反向弧
    int head[maxn],next[maxm];  //邻接表表头和next数组
    bool vis[maxn];             //BFS使用,标记一个节点是否被遍历过
    int d[maxn];                //从起点到i点的距离
    int cur[maxn];              //当前弧下标

    void init(int n,int s,int t)
    {
        this->n=n,this->s=s,this->t=t;
        memset(head,-1,sizeof(head));
        m=0;
    }

    void AddEdge(int from,int to,int cap)
    {
        edges[m]= Edge(from,to,cap,0) ;
        next[m]=head[from];
        head[from]=m++;

        edges[m]= Edge(to,from,0,0) ;
        next[m]=head[to];
        head[to]=m++;
    }

    bool BFS()
    {
        memset(vis,0,sizeof(vis));
        queue<int> Q;//用来保存节点编号的
        Q.push(s);
        d[s]=0;
        vis[s]=true;
        while(!Q.empty())
        {
            int x=Q.front(); Q.pop();
            for(int i=head[x]; i!=-1; i=next[i])
            {
                Edge& e=edges[i];
                if(!vis[e.to] && e.cap>e.flow)
                {
                    vis[e.to]=true;
                    d[e.to] = d[x]+1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x,int a)
    {
        if(x==t || a==0)return a;
        int flow=0,f;//flow用来记录从x到t的最小残量
        for(int& i=cur[x]; i!=-1; i=next[i])
        {
            Edge& e=edges[i];
            if(d[x]+1==d[e.to] && (f=DFS( e.to,min(a,e.cap-e.flow) ) )>0 )
            {
                e.flow +=f;
                edges[i^1].flow -=f;
                flow += f;
                a -= f;
                if(a==0) break;
            }
        }
        return flow;
    }

    int Maxflow()
    {
        int flow=0;
        while(BFS())
        {
            for(int i=1;i<=n;i++) cur[i]=head[i];
            flow += DFS(s,inf);
        }
        return flow;
    }
}DC;

int main()
{
    int T; scanf("%d",&T);
    for(int kase=1; kase<=T; ++kase)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        DC.init(n,1,n);
        while(m--)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            DC.AddEdge(u,v,w);
        }
        printf("Case %d: %d
",kase,DC.Maxflow());
    }
    return 0;
}
原文地址:https://www.cnblogs.com/lalalatianlalu/p/9033427.html