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 99999999

struct 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;
}