POJ 2289 Jamie's Contact Groups

二分答案+网络最大流

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

int N,M;
const int maxn = 2000 + 10;
const int INF = 0x7FFFFFFF;
struct Edge
{
    int from, to, cap, flow;
    Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f){}
};
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
int n, m, s, t;
int U[maxn*500],V[maxn*500];
int tot;
int T[maxn];
char ss[maxn];

void init()
{
    for(int i = 0; i < maxn; i++) G[i].clear();
    edges.clear();
    tot=0;
    memset(T,0,sizeof T);
}

void AddEdge(int from, int to, int cap)
{
    edges.push_back(Edge(from, to, cap, 0));
    edges.push_back(Edge(to, from, 0, 0));
    int w = edges.size();
    G[from].push_back(w - 2);
    G[to].push_back(w - 1);
}

bool BFS()
{
    memset(vis, 0, sizeof(vis));
    queue<int>Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty())
    {
        int x = Q.front();
        Q.pop();
        for (int i = 0; i<G[x].size(); i++)
        {
            Edge e = edges[G[x][i]];
            if (!vis[e.to] && e.cap>e.flow)
            {
                vis[e.to] = 1;
                d[e.to] = d[x] + 1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}

int DFS(int x, int a)
{
    if (x == t || a == 0)
        return a;
    int flow = 0, f;
    for (int &i = cur[x]; i<G[x].size(); i++)
    {
        Edge e = edges[G[x][i]];
        if (d[x]+1 == d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0)
        {
            edges[G[x][i]].flow+=f;
            edges[G[x][i] ^ 1].flow-=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    if(!flow) d[x] = -1;
    return flow;
}

int dinic(int s, int t)
{
    int flow = 0;
    while (BFS())
    {
        memset(cur, 0, sizeof(cur));
        flow += DFS(s, INF);
    }
    return flow;
}

void Input()
{
    int SS;
    for(int i=1;i<=N;i++)
    {
        gets(ss);
        for(int ii=0;ss[ii];ii++)
            if(ss[ii]==' ')
                {SS=ii;break;}
        int num=0;
        for(int ii=SS+1;;ii++)
        {
            if(ss[ii]>='0'&&ss[ii]<='9') num=num*10+ss[ii]-'0';
            if(ss[ii]==' '||ss[ii]=='\0')
            {
                U[tot]=i;V[tot]=N+num+1;tot++;
                T[num]++;
                num=0;
            }
            if(ss[ii]=='\0') break;
        }
    }
}

void Solve()
{
    int Ma=-INF;
    for(int i=0;i<M;i++) if(T[i]>Ma) Ma=T[i];
    int Min=0,Max=Ma;
    int Mid=(Min+Max)/2;
    s=0;t=N+M+1;
    while(1)
    {
        for(int i = 0; i < maxn; i++) G[i].clear();
        edges.clear();
        for(int i=1;i<=N;i++) AddEdge(s,i,1);
        for(int i=0;i<tot;i++) AddEdge(U[i],V[i],INF);
        for(int i=N+1;i<=N+M;i++) AddEdge(i,t,Mid);
        if(dinic(s,t)<N)
        {
            Min=Mid+1;
            Mid=(Min+Max)/2;
        }
        else
        {
            Max=Mid;
            Mid=(Min+Max)/2;
        }
        if(Min==Max) break;
    }
    printf("%d\n",Min);
}

int main()
{
    while(~scanf("%d%d",&N,&M))
    {
        if(!N&&!M) break;
        scanf("\n");
        init();
        Input();
        Solve();
    }
    return 0;
}

 

posted @ 2015-09-20 10:59  Fighting_Heart  阅读(184)  评论(0编辑  收藏  举报