【网络流最大流】poj3281Dining

来源：互联网发布：王兆山真相知乎编辑：程序博客网时间：2024/04/30 17:51
/*EK算法版本的，比较慢哦。。。。。详见下面dinic版本-----------------------------------------题目是网络流最大流的问题----------------------------------------建图：关键：拆点把每个牛拆成两个点，牛作为一个点有流量限制，即每一头牛只能的一份饭。把牛拆开后，将边的权值赋值为1，----------------------------------------建好图后就可以EK算法最大流了----------------------------------------建图代码：for(int i=1; i<=f; i++)源点和food相连    g[s][i] = 1;for(int i=1; i<=d; i++)drink和汇点相连    g[i + 2*n + f][t] = 1;for(int i=1; i<=n; i++){    cin>>a>>b;    for(int j=1; j<=a; j++)    {        cin>>food;        g[food][f + i] = 1;food和牛相连    }    g[f + i][f + i + n] = 1;牛拆点后也要相连    for(int j=1; j<=b; j++)    {        cin>>drink;        g[f + i + n][2*n + f + drink] = 1;牛和drink相连    }}---------------------------------*/#include <iostream>#include <cstdio>#include <queue>#include <cstring>#define INF 0x3f3f3f3fusing namespace std;const int N = 500;int n,f,d;int g[N][N];int flow[N][N],a[N],p[N];int s,t;int EK(int s, int t){    queue<int> q;    int f = 0;    for(;;)    {        memset(a,0,sizeof(a));        a[s] = INF;        q.push(s);        while(!q.empty())        {            int u = q.front();            q.pop();            for(int v=s; v<=t; v++)            {                if(!a[v] && g[u][v] > flow[u][v])                {                    p[v] = u;                    q.push(v);                    a[v] = min(a[u],g[u][v] - flow[u][v]);                }            }        }        if(a[t] == 0)            break;        for(int u=t; u != s; u=p[u])        {            flow[p[u]][u] += a[t];            flow[u][p[u]] -= a[t];        }        f += a[t];    }    return f;}void init(){    int a,b,food,drink;    s = 0;    t = 2*n + f + d + 1;    memset(g,0,sizeof(g));    for(int i=1; i<=f; i++)        g[s][i] = 1;    for(int i=1; i<=d; i++)        g[i + 2*n + f][t] = 1;    for(int i=1; i<=n; i++)    {        cin>>a>>b;        for(int j=1; j<=a; j++)        {            cin>>food;            g[food][f + i] = 1;        }        g[f + i][f + i + n] = 1;        for(int j=1; j<=b; j++)        {            cin>>drink;            g[f + i + n][2*n + f + drink] = 1;        }    }}int main(){    //freopen("input.txt","r",stdin);    while(scanf("%d%d%d",&n,&f,&d) != EOF)    {        init();        printf("%d\n",EK(s, t));    }    return 0;}
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战斗，毫不退缩；奋斗，永不停歇~~~~~~~~~~~~~~
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