POJ 1274 The Perfect Stall

The Perfect Stall

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 6

Problem Description

Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall. 
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible. 

 

Input

The input includes several cases. For each case, the first line contains two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. Each of the following N lines corresponds to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow.

 

Output

For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made.

 

Sample Input

5 52 2 53 2 3 42 1 53 1 2 51 2 

 

Sample Output

4

 

网络流或者二分匹配。

题意：  有n头牛  m个摊位  每个摊位只能容纳一头奶牛 每头牛都有自己愿意产奶的摊位   问最多有几头牛能在愿意的摊位产奶

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;#define MAXN 44444#define MAXM 999999#define inf 1<<30struct Edge{    int v,cap,next;} edge[MAXM];int n,m,vs,vt,NE,NV;int head[MAXN];void Insert(int u,int v,int cap){    edge[NE].v=v;    edge[NE].cap=cap;    edge[NE].next=head[u];    head[u]=NE++;    edge[NE].v=u;    edge[NE].cap=0;    edge[NE].next=head[v];    head[v]=NE++;}int level[MAXN];int gap[MAXN];void bfs(int vt){    memset(level,-1,sizeof(level));    memset(gap,0,sizeof(gap));    level[vt]=0;    gap[level[vt]]++;    queue<int>que;    que.push(vt);    while(!que.empty())    {        int u=que.front();        que.pop();        for(int i=head[u]; i!=-1; i=edge[i].next)        {            int v=edge[i].v;            if(level[v]!=-1)continue;            level[v]=level[u]+1;            gap[level[v]]++;            que.push(v);        }    }}int pre[MAXN];int cur[MAXN];//参数  起点  终点int SAP(int vs,int vt){    bfs(vt);    memset(pre,-1,sizeof(pre));    memcpy(cur,head,sizeof(head));    int u=pre[vs]=vs,flow=0,aug=inf;    gap[0]=NV;    while(level[vs]<NV)    {        bool flag=false;        for(int &i=cur[u]; i!=-1; i=edge[i].next)        {            int v=edge[i].v;            if(edge[i].cap&&level[u]==level[v]+1)            {                flag=true;                pre[v]=u;                u=v;                //  aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));                aug=min(aug,edge[i].cap);                if(v==vt)                {                    flow+=aug;                    for(u=pre[v]; v!=vs; v=u,u=pre[u])                    {                        edge[cur[u]].cap-=aug;                        edge[cur[u]^1].cap+=aug;                    }                    //     aug=-1;                    aug=inf;                }                break;            }        }        if(flag)continue;        int minlevel=NV;        for(int i=head[u]; i!=-1; i=edge[i].next)        {            int v=edge[i].v;            if(edge[i].cap&&level[v]<minlevel)            {                minlevel=level[v];                cur[u]=i;            }        }        if(--gap[level[u]]==0)break;        level[u]=minlevel+1;        gap[level[u]]++;        u=pre[u];    }    return flow;}int main(){    int u,v,w,w1,w2;    while(~scanf("%d%d",&n,&m))    {        vs=0;//起点        vt=n+m+1;//终点        NV=n+m+1;//可能走过的边的总数        NE=0;        memset(head,-1,sizeof(head));        for(int i=1; i<=n; i++)        {            //scanf("%d%d",&w1,&w2);            Insert(0,i,1);            //Insert(i,vt,w2);        }        for(int i=1; i<=m; i++)        {            //scanf("%d%d%d",&u,&v,&w);            Insert(i+n,n+m+1,1);            //Insert(v,u,w);        }        for(int i=1;i<=n;i++)        {            scanf("%d",&u);            for(int j=0;j<u;j++)            {                scanf("%d",&v);                Insert(i,v+n,1);            }        }        printf("%d\n",SAP(vs,vt));    }    return 0;}