POJ 3281 网络流（dinic邻接矩阵、单路增广、多路增广）

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思路：刚开始看题就想到怎么建图了，源点连向所有的食物，食物连牛，牛连饮料，饮料连汇点，所有的流量都是1.不过这样建图好后，WA了。原来是一头牛只能单一匹配一组食物和饮料，所以牛得拆点，牛之间得相连，流量为1，以保证单一匹配食物和饮料。

邻接矩阵dinic单路的代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<cmath>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define llson j<<1,l,mid#define rrson j<<1|1,mid+1,r#define INF 0x7fffffff#define maxn 20005typedef long long ll;typedef unsigned long long ull;using namespace std;int n,m,d[maxn],Map[501][501];int bfs(){    mem(d,-1);    queue<int>q; q.push(0); d[0]=0;    while(!q.empty())  //找增广路    {        int u=q.front(); q.pop();        for(int v=0;v<=n;v++)        {            if(d[v]==-1&&Map[u][v])            {                d[v]=d[u]+1; //分层，越往后，层数越大                q.push(v);            }        }    }    return d[n]!=-1;}int dfs(int u,int Min){    int sum;    if(u==n) return Min;    for(int v=0;v<=n;v++)        if(Map[u][v]&&d[v]==d[u]+1&&(sum=dfs(v,min(Min,Map[u][v]))))        {            Map[u][v]-=sum;            Map[v][u]+=sum;            return sum;        }    return 0;}int Dinic(){    int tmp,ans=0;    while(bfs())    {        while(tmp=dfs(0,INF))            ans+=tmp;    }    return ans;}int main(){    //freopen("1.txt","r",stdin);    int N,F,D,ff,dd,i,j,mm;    scanf("%d%d%d",&N,&F,&D);    for(i=1;i<=N;i++)    {        scanf("%d%d",&ff,&dd);        for(j=0;j<ff;j++)        {            scanf("%d",&mm);            Map[mm+N*2][i]=1;        }        for(j=0;j<dd;j++)        {            scanf("%d",&mm);            Map[i+N][mm+N*2+F]=1;        }        Map[i][i+N]=1;//牛之间自连，保存每头牛只匹配一组食物和饮料    }    for(i=N*2+1;i<=N*2+F;i++)        Map[0][i]=1;//连源点    n=N*2+F+D+1;    for(i=N*2+F+1;i<=N*2+F+D;i++)        Map[i][n]=1;//连汇点    printf("%d\n",Dinic());    return 0;}

dinic单路增广代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<cmath>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define llson j<<1,l,mid#define rrson j<<1|1,mid+1,r#define INF 0x7ffffffftypedef long long ll;typedef unsigned long long ull;using namespace std;#define maxn 2000005int n,cnt,d[maxn],head[maxn],go[maxn],q[maxn];struct node{    int v,w,next;}e[maxn];void add(int u,int v,int w){    e[cnt].v=v; e[cnt].w=w; go[cnt]=cnt+1;    e[cnt].next=head[u]; head[u]=cnt++;    e[cnt].v=u; e[cnt].w=0; go[cnt]=cnt-1;//刚开始反向边权值写成w了一直wa    e[cnt].next=head[v]; head[v]=cnt++;}int bfs(){    mem(d,-1);    int l=0,r=0;    q[r++]=0; d[0]=0;    while(l<r)  //找增广路    {        int u=q[l++];        for(int v=head[u];v!=-1;v=e[v].next)        {            if(d[e[v].v]==-1&&e[v].w)            {                d[e[v].v]=d[u]+1; //分层，越往后，层数越大                if(e[v].v!=n+1) q[r++]=e[v].v;            }        }    }    return d[n]!=-1;}int dfs(int u,int Min){    int sum;    if(u==n) return Min;    for(int v=head[u];v!=-1;v=e[v].next)  //Min-duolu这个多路增广的做法真神！        if(e[v].w&&d[e[v].v]==d[u]+1&&(sum=dfs(e[v].v,min(Min,e[v].w))))        {            e[v].w-=sum;            e[go[v]].w+=sum;            return sum;        }    return 0;}void init(){    mem(head,-1),mem(go,0),cnt=0;}int Dinic(){    int tmp,ans=0;    while(bfs())    {        while(tmp=dfs(0,INF)) ans+=tmp;    }    return ans;}int main(){    //freopen("1.txt","r",stdin);    int N,F,D,ff,dd,i,j,mm;    scanf("%d%d%d",&N,&F,&D);    init();    for(i=1;i<=N;i++)    {        scanf("%d%d",&ff,&dd);        for(j=0;j<ff;j++)        {            scanf("%d",&mm);            add(mm+N*2,i,1);        }        for(j=0;j<dd;j++)        {            scanf("%d",&mm);            add(i+N,mm+N*2+F,1);        }        add(i,i+N,1);//牛之间自连，保存每头牛只匹配一组食物和饮料    }    for(i=N*2+1;i<=N*2+F;i++)        add(0,i,1);//连源点    n=N*2+F+D+1;    for(i=N*2+F+1;i<=N*2+F+D;i++)        add(i,n,1);//连汇点    printf("%d\n",Dinic());    return 0;}

Dinic多路增广代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<set>#include<cmath>#include<bitset>#define mem(a,b) memset(a,b,sizeof(a))#define lson i<<1,l,mid#define rson i<<1|1,mid+1,r#define llson j<<1,l,mid#define rrson j<<1|1,mid+1,r#define INF 0x7ffffffftypedef long long ll;typedef unsigned long long ull;using namespace std;#define maxn 2000005int n,cnt,d[maxn],head[maxn],go[maxn],q[maxn];struct node{    int v,w,next;}e[maxn];void add(int u,int v,int w){    e[cnt].v=v; e[cnt].w=w; go[cnt]=cnt+1;    e[cnt].next=head[u]; head[u]=cnt++;    e[cnt].v=u; e[cnt].w=0; go[cnt]=cnt-1;//刚开始反向边权值写成w了一直wa    e[cnt].next=head[v]; head[v]=cnt++;}int bfs(){    mem(d,-1);    int l=0,r=0;    q[r++]=0; d[0]=0;    while(l<r)  //找增广路    {        int u=q[l++];        for(int v=head[u];v!=-1;v=e[v].next)        {            if(d[e[v].v]==-1&&e[v].w)            {                d[e[v].v]=d[u]+1; //分层，越往后，层数越大                if(e[v].v!=n+1) q[r++]=e[v].v;            }        }    }    return d[n]!=-1;}int dfs(int u,int Min){    int sum,duolu=0;    if(u==n) return Min;    for(int v=head[u];v!=-1&&Min-duolu>0;v=e[v].next)  //Min-duolu这个多路增广的做法真神！        if(e[v].w&&d[e[v].v]==d[u]+1&&(sum=dfs(e[v].v,min(Min-duolu,e[v].w))))        {            e[v].w-=sum;            e[go[v]].w+=sum;            duolu+=sum;        }    return duolu;}void init(){    mem(head,-1),mem(go,0),cnt=0;}int Dinic(){    int tmp,ans=0;    while(bfs())    {        while(tmp=dfs(0,INF)) ans+=tmp;    }    return ans;}int main(){    //freopen("1.txt","r",stdin);    int N,F,D,ff,dd,i,j,mm;    scanf("%d%d%d",&N,&F,&D);    init();    for(i=1;i<=N;i++)    {        scanf("%d%d",&ff,&dd);        for(j=0;j<ff;j++)        {            scanf("%d",&mm);            add(mm+N*2,i,1);        }        for(j=0;j<dd;j++)        {            scanf("%d",&mm);            add(i+N,mm+N*2+F,1);        }        add(i,i+N,1);//牛之间自连，保存每头牛只匹配一组食物和饮料    }    for(i=N*2+1;i<=N*2+F;i++)        add(0,i,1);//连源点    n=N*2+F+D+1;    for(i=N*2+F+1;i<=N*2+F+D;i++)        add(i,n,1);//连汇点    printf("%d\n",Dinic());    return 0;}

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