Jamie's Contact Groups POJ - 2289

来源：互联网发布：设计师笔记本2017知乎编辑：程序博客网时间：2024/06/02 00:51

题目连接点这里

二分图多重匹配模版题，

二分答案后，跑匈牙利就可以了

#include<iostream>#include<stdio.h>#include<queue>#include<algorithm>#include<string.h>#define MX 2222#define INF 0x3f3f3f3f#define mem(x,y) memset(x,y,sizeof(x))#define FIN freopen("input.txt","r",stdin)using namespace std;int n,m;int head[MX],rear;int mid;struct{    int to,nxt;} edge[1111*MX];void edge_init(){    mem(head,-1);    rear=0;}void edge_add(int a,int b){    edge[rear].to=b;    edge[rear].nxt=head[a];    head[a]=rear++;}bool vis[MX];int match[MX][1111];int mcnt[MX];bool hungarian_dfs(int u){    for(int i=head[u]; ~i; i=edge[i].nxt)    {        int v=edge[i].to;        if(vis[v]) continue;        vis[v]=1;        if(mcnt[v]<mid)        {            match[v][++mcnt[v]]=u;            return 1;        }        for(int j=1; j<=mid; j++)            if(hungarian_dfs(match[v][j]))            {                match[v][j]=u;                return 1;            }    }    return 0;}int hungarian(){    mem(mcnt,0);    int ret=0;    for(int i=1; i<=n; i++)    {        mem(vis,0);        if(hungarian_dfs(i)==0) return 0;    }    return 1;}int erfeng(int l,int r){    while(l<r)    {        mid=((l+r)>>1);        if(hungarian()) r=mid;        else l=mid+1;    }    return l;}bool lof;bool Int(int &a){    if(lof) return lof=0;    char c;    for(c=getchar(); c<'0'||c>'9'; c=getchar())if(c=='\n')return 0;    a=c-'0';    for(c=getchar(); c>='0'&&c<='9'; c=getchar())a=a*10+c-'0';    lof=c=='\n';    return 1;}int main(){    FIN;    while(scanf("%d%d",&n,&m)&&n)    {        edge_init();        char s[1111];        for(int i=1; i<=n; i++)        {            scanf("%s",s);            lof=0;            int x;            while(Int(x)) edge_add(i,x+n+1),edge_add(x+n+1,i);        }        printf("%d\n",erfeng(1,n));    }    return 0;}

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