【poj3281】【最大流】Dining

一道最大流的简单题。将食物和饮料分别看成一个点集，食物放在左边(记为f[i])，饮料放在右边(记为d[i])。然后将每头牛拆成两个点k,k'

然后开始建图：

1、对于每头牛喜欢的食物和饮料:添加edge(f[i],k,1),edge(d[i],k',1),edge(k,k',1)，这样做的目的是为了限制每头牛只能享用一种食物和饮料。

2、对于每个食物和饮料:添加edge(s,f[i],1),edge(d[i],t,1),这样做的目的是限制一种食物和饮料只能选一次。

这样在原图上做一次最大流即为所求，要注意的是拆点时一定要注意好点的映射关系，不能重复映射。

代码：

#include<cstdio>
#include<cstring>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 500;
const int maxm = 50000;
struct Edge
{
int c,pos;
int next;
}E[maxm];
int dis[maxn],gap[maxn],pre[maxn],cur[maxn];
int head[maxn];
int NE = 0;
int n,f,d,add;
int s,t,nodenum;
void init()
{
freopen("poj3281.in","r",stdin);
freopen("poj3281.out","w",stdout);
}

void insert(int u,int v,int c)
{
E[NE].c = c;E[NE].pos = v;
E[NE].next = head[u];head[u] = NE++;
E[NE].c = 0;E[NE].pos = u;
E[NE].next = head[v];head[v] = NE++;
}

inline void checkmin(int &a,int b)
{
if(a == -1 || a > b)a = b;
}

void readdata()
{
memset(head,-1,sizeof(head));
scanf("%d%d%d",&n,&f,&d);
add = (n << 1);
for(int i = 1;i <= n;i++)
{
int fo,dr;
scanf("%d%d",&fo,&dr);
for(int j = 1;j <= fo;j++)
{
int tmp;
scanf("%d",&tmp);
insert(tmp + add,i,1);
}
for(int j = 1;j <= dr;j++)
{
int tmp;
scanf("%d",&tmp);
insert(i + n,tmp + add + f,1);
}
insert(i,i + n,1);
}
}

int sap()
{
memset(dis,0,sizeof(dis));
memset(gap,0,sizeof(gap));
for(int i = s;i <= t;i++)cur[i] = head[i];
int u = pre[s] = s,maxflow = 0,aug = -1;
gap[0] = nodenum;
while(dis[s] < nodenum)
{
loop:   for(int &i = cur[u];i != -1;i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && dis[u] == dis[v] + 1)
{
checkmin(aug,E[i].c);
pre[v] = u;
u = v;
if(v == t)
{
maxflow += aug;
for(u = pre[u];v != s;v = u,u = pre[u])
{
E[cur[u]].c -= aug;
E[cur[u]^1].c += aug;
}
aug = -1;
}
goto loop;
}
}
int mind = nodenum;
for(int i = head[u];i != -1;i = E[i].next)
{
int v = E[i].pos;
if(E[i].c && (mind > dis[v]))
{
cur[u] = i;
mind = dis[v];
}
}
if((--gap[dis[u]]) == 0)break;
gap[dis[u] = mind + 1]++;
u = pre[u];
}
return maxflow;
}

void solve()
{
s = 0,t = n * 2 + f + d + 1;
nodenum = t + 1;
for(int i = 1;i <= f;i++)insert(s,i + add,1);
for(int i = 1;i <= d;i++)insert(i + add + f,t,1);
printf("%d\n",sap());
}

int main()
{
init();
readdata();
solve();
return 0;
}