POJ1149_PIGS_网络流

#include <iostream>#include <string.h>#include <cstdio>#include <vector>#include <queue>#define INF 0x3f3f3f3f#define MAXM 1010#define MAXN 110using namespace std;int cap[MAXM];struct Edge{    Edge(int from, int to, int cap, int flow): from(from), to(to), cap(cap), flow(flow) {}    int from, to, cap, flow;};struct Dinic{    int n,m,s,t;            //结点数,边数(包括反向弧),源点与汇点编号    vector<Edge> edges;     //边表 edges[e]和edges[e^1]互为反向弧    vector<int> G[MAXN];    //邻接表,G[i][j]表示结点i的第j条边在e数组中的序号    bool vis[MAXN];         //BFS使用,标记一个节点是否被遍历过    int d[MAXN];            //从起点到i点的距离    int cur[MAXN];          //当前弧下标    void init(int n,int s,int t)    {        this->n = n, this->s = s,this->t = t;        for(int i = 0; i < n; i++)            G[i].clear();        edges.clear();    }    void AddEdge(int from,int to,int cap)    {        edges.push_back(Edge(from, to, cap, 0));        edges.push_back(Edge(to, from, 0, 0));        m = edges.size();        G[from].push_back(m - 2);        G[to].push_back(m - 1);    }    bool BFS()    {        memset(vis, 0, sizeof(vis));        queue<int> Q;//用来保存节点编号的        Q.push(s);        d[s] = 0;        vis[s] = true;        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] = true;                    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;//flow用来记录从x到t的最小残量        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)            {                e.flow += f;                edges[G[x][i]^1].flow -= f;                flow += f;                a -= f;                if(a == 0)                    break;            }        }        return flow;    }    int Maxflow()    {        int flow = 0;        while(BFS())        {            memset(cur, 0, sizeof(cur));            flow += DFS(s, INF);        }        return flow;    }} DC;int book[MAXM], book1[MAXN];int main(){    //freopen("aa.txt", "r", stdin);    int m, n, a, b;    memset(book, 0, sizeof(book));    memset(book1, 0, sizeof(book1));    DC.init(MAXN, 0, MAXN - 1);    scanf("%d %d", &m, &n);    for (int i = 1; i <= m; i++)        scanf("%d", &cap[i]);    int key;    for (int i = 1; i <= n; i++)  //n  customer   m house    {        cin >> a;        while (a--)        {            cin >> key;            if (!book[key])            {                if (!book1[i])                {                    DC.AddEdge(0, i, cap[key]);                    book1[i] = DC.edges.size() - 2;                }                else                    DC.edges[book1[i]].cap += cap[key];                book[key] = i;            }            else                DC.AddEdge(book[key], i, INF);        }        cin >> b;        DC.AddEdge(i, MAXN - 1, b);    }    cout << DC.Maxflow() << endl;}

题目大意
若干个顾客依次来到猪圈，各自有若干个猪圈的钥匙，各自有做多买猪量，每个猪圈有若干头猪，每个顾客到达并打开他有钥匙的猪圈买走猪后，能交将剩下的猪在打开的猪圈中人以分配。问最多能卖出多少猪。
思路
建立超级源点和超级汇点，顾客和源点的边权值为他拥有钥匙的猪圈的猪总数，和汇点的边权值为他最多能购买的猪的数量。顾客之间能建立一种关系：a顾客先于顾客b到达某猪圈，则a和b之间可以建立一条容量为无穷大的边。跑一遍网络流。