网络流( ISAP + 拆点 )——Dining ( POJ 3281 )

• 题目链接：
  http://poj.org/problem?id=3281

• 分析：
  一头牛只吃指定的几种食物或饮料，但是每一种食物和饮料都只有一个，给出N头牛及其要求，F种食物和D种饮料，问最多可以满足多少头牛的需要（饮料和食物都满足需求）。

• 题解：
  1.建图：需要将每头牛拆成牛1和牛2，然后建立一个源点S，从S出发作边连接每一个食物，容量为1。然后从食物出发向和其对应的牛做边，容量为1。再连接牛1和牛2，容量为1。从牛2出发向与其对应的饮料作边，容量为1。再连接饮料与汇点，容量为1。

//邻接链表存储边int head[Maxn];struct node{    int v, cap; //v为该点连的当前边另一端的点，cap为当前边边的容量    int next;//该点连接的下一条边}edge[80000];void add(int u, int v, int cap)//添加边{    edge[cont].v = v;    edge[cont].cap = cap;    edge[cont].next = head[u];    head[u] = cont++;    edge[cont].v=u; //反向建边，容量为0    edge[cont].cap=0;    edge[cont].next=head[v];    head[v] = cont++;}

2.BFS：

int dis[Maxn];int num[Maxn];void BFS(int source,int sink){    queue<int>q;    while(！q.empty())        q.pop();    memset(num,0,sizeof(num));    memset(dis,-1,sizeof(dis));    q.push(sink);    dis[sink]=0;    num[0]=1;    while(!q.empty())    {        int u=q.front();        q.pop();        for(int i = head[u]; i! = -1; i = edge[i].next)        {            int v = edge[i].v;            if(dis[v] == -1)            {                dis[v] = dis[u] + 1;                num[dis[v]]++;                q.push(v);            }        }    }}

3.ISAP：

int cur[Maxn];int pre[Maxn];int ISAP(int source,int sink,int n){    memcpy(cur,head,sizeof(cur));    int flow=0, u = pre[source] = source;    BFS(source, sink);    while( dis[source] < n )    {        if(u == sink)        {            int df = INF, pos;            for(int i =source;i != sink;i = edge[cur[i]].v)            {                if(df > edge[cur[i]].cap)                {                    df = edge[cur[i]].cap;                    pos = i;                }            }            for(int i = source;i != sink;i = edge[cur[i]].v)            {                edge[cur[i]].cap -= df;                edge[cur[i]^1].cap += df;            }            flow += df;            u = pos;        }        int st;        for(st = cur[u];st != -1;st = edge[st].next)        {            if(dis[edge[st].v] + 1 == dis[u] && edge[st].cap)            {                break;            }        }        if(st != -1)        {            cur[u] = st;            pre[edge[st].v] = u;            u = edge[st].v;        }        else        {            if( (--num[dis[u]])==0 ) break;            int mind = n;            for(int id = head[u];id != -1;id = edge[id].next)            {                if(mind > dis[edge[id].v] && edge[id].cap != 0)                {                    cur[u] = id;                    mind = dis[edge[id].v];                }            }            dis[u] = mind+1;            num[dis[u]]++;            if(u!=source)            u = pre[u];        }    }    return flow;}

• AC代码：

#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;#define Maxn 2000#define INF 99999999struct node{    int v, cap;    int next;}edge[80000];int N,F,D;int cont;int head[Maxn];int dis[Maxn];int num[Maxn];int cur[Maxn];int pre[Maxn];void add(int u, int v, int cap){    edge[cont].v = v;    edge[cont].cap = cap;    edge[cont].next = head[u];    head[u] = cont++;    edge[cont].v=u;    edge[cont].cap=0;    edge[cont].next=head[v];    head[v] = cont++;}void BFS(int source,int sink){    queue<int>q;    while(q.empty()==false)    q.pop();    memset(num,0,sizeof(num));    memset(dis,-1,sizeof(dis));    q.push(sink);    dis[sink]=0;    num[0]=1;    while(!q.empty())    {        int u=q.front();        q.pop();        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v = edge[i].v;            if(dis[v] == -1)            {                dis[v] = dis[u] + 1;                num[dis[v]]++;                q.push(v);            }        }    }}int ISAP(int source,int sink,int n){    memcpy(cur,head,sizeof(cur));    int flow=0, u = pre[source] = source;    BFS( source,sink);    while( dis[source] < n )    {        if(u == sink)        {            int df = INF, pos;            for(int i =source;i != sink;i = edge[cur[i]].v)            {                if(df > edge[cur[i]].cap)                {                    df = edge[cur[i]].cap;                    pos = i;                }            }            for(int i = source;i != sink;i = edge[cur[i]].v)            {                edge[cur[i]].cap -= df;                edge[cur[i]^1].cap += df;            }            flow += df;            u = pos;        }        int st;        for(st = cur[u];st != -1;st = edge[st].next)        {            if(dis[edge[st].v] + 1 == dis[u] && edge[st].cap)            {                break;            }        }        if(st != -1)        {            cur[u] = st;            pre[edge[st].v] = u;            u = edge[st].v;        }        else        {            if( (--num[dis[u]])==0 ) break;            int mind = n;            for(int id = head[u];id != -1;id = edge[id].next)            {                if(mind > dis[edge[id].v] && edge[id].cap != 0)                {                    cur[u] = id;                    mind = dis[edge[id].v];                }            }            dis[u] = mind+1;            num[dis[u]]++;            if(u!=source)            u = pre[u];        }    }    return flow;}void init(){       memset(head,-1,sizeof(head));       cont=0;}int main(){    int a,b,c;    while(~scanf("%d%d%d", &N, &F, &D))    {        init();        int S = 0;        int T = 2*N+F+D+1;        for(int i = 1; i<=F; i++) //建立源点到所有食物        {            add(S, i, 1);        }        for(int i = 1; i<=N; i++)        {            scanf("%d%d",&a,&b);            for(int j = 1;j<=a;j++)            {                scanf("%d",&c);                add(c,F+i,1);//食物到牛            }            for(int j = 1;j<=b;j++)            {                scanf("%d",&c);                add(F+i+N,2*N+F+c,1);//牛' 到饮料            }                add(F+i,F+i+N,1);//牛 到 牛’        }        for(int i = 1;i<=D;i++)        {            add(2*N+F+i, T,1);//食物 到汇点        }        int ans = ISAP(S, T, T+1 );        printf("%d\n",ans);    }    return 0;}