POJ1273 Drainage Ditches 最大流模板题（dinic）

最大流的模板题

给出边数M，顶点数N 以及每条边的容量 求1到N的最大流

注意可以有重边

邻接矩阵模板：

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxx 0x3f3f3f
#define M   205
using namespace std;
int arc[M][M];   //弧的剩余流量
int level[M];
int n;
int min(int a,int b)
{
    return a<b?a:b;
}
int bfs()
{
    int q[M];
    int head=0,tail=0,v;
    for(int i=1; i<=n; i++)
        level[i]=-1;
    level[1]=0;
    q[tail++]=1;
    while(head<tail)
    {
        v=q[head++];
        for(int i=1; i<=n; i++)
            if(level[i]==-1&&arc[v][i])
            {
                level[i]=level[v]+1;
                q[tail++]=i;
            }
    }
    return level[n]!=-1;
}

int dfs(int loc,int flow)
{
    int unused=flow;
    if(loc==n||flow==0)
        return flow;
    for(int i=1; i<=n; i++)
        if(arc[loc][i]&&level[loc]+1==level[i])
        {
            int used=dfs(i,min(unused,arc[loc][i]));
            arc[loc][i]-=used;
            arc[i][loc]+=used;
            unused-=used;
        }
    return flow-unused;
}

int main()
{
    int m;
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=i; j++)
                arc[i][j]=arc[j][i]=0;
        int a,b,f;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&f);
            arc[a][b]+=f;
        }
        int ans=0;
        while(bfs())
            ans+=dfs(1,maxx);

        cout<<ans<<endl;
    }
    return 0;
}

邻接表模板：

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxx 0x3f3f3f
#define M   205
using namespace std;
struct arc{
    int to,value;
    int pre;
}edge[45555];
int level[M];
int table[M];
int n;
int sum;
int min(int a,int b)
{
    return a<b?a:b;
}
void addedge(int a,int b,int v)
{
    edge[sum]={b,v,table[a]};
    table[a]=sum++;
    edge[sum]={a,0,table[b]};
    table[b]=sum++;
    return;
}
int bfs()
{
    int q[M];
    int head=0,tail=0,v;
    for(int i=1; i<=n; i++)
        level[i]=-1;
    level[1]=0;
    q[tail++]=1;
    while(head<tail)
    {
        v=q[head++];
        for(int i=table[v]; i!=-1; i=edge[i].pre)
            if(level[edge[i].to]==-1&&edge[i].value)
            {
                level[edge[i].to]=level[v]+1;
                q[tail++]=edge[i].to;
            }
    }
    return level[n]!=-1;
}
int ss=0;
int dfs(int loc,int flow)
{
    int unused=flow;
    if(loc==n||flow==0)
        return flow;
    for(int i=table[loc]; i!=-1; i=edge[i].pre)
        if(edge[i].value&&level[loc]+1==level[edge[i].to])
        {
            int used=dfs(edge[i].to,min(unused,edge[i].value));
            edge[i].value-=used;
            edge[i^1].value+=used; //正向边反向边的的序号的奇偶性不同且差一
            unused-=used;
        }
    return flow-unused;
}

int main()
{
    int m;
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=1;i<=n;i++)
            table[i]=-1;
        sum=0;
        int a,b,f;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&f);
            addedge(a,b,f);
        }
        int ans=0;
        while(bfs())
            ans+=dfs(1,maxx);

        cout<<ans<<endl;
    }
    return 0;
}