POJ 1459最大流模板

本文详细介绍了一种基于网络流算法的实现方法,通过构建超级源点和汇点来解决实际问题。文章提供了完整的C++代码示例,包括节点添加、宽度优先搜索(BFS)、深度优先搜索(DFS)等关键步骤,并展示了如何利用Dinic算法求解最大流。

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题意大家自己去网上找吧。。。
要自己建立超级源点,汇点
本弱渣直到现在才正式敢于面对网络流,,,
只是想写成自己熟悉的模板而已


//POJ1459
//Memory:1060K
//Time: 797MS
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;

const int MAXN = 110;
const int INF = 1e9+9;

struct Node
{
    int to; ///终点
    int cap; ///容量
    int rev; ///反向边
    Node(int _to,int _cap,int _rev):to(_to),cap(_cap),rev(_rev){}

};

vector <Node> edge[MAXN];

void add_node(int from,int to,int cap)
{
   // v[from].push_back((Node){to,cap,v[to].size()});
    edge[from].push_back(Node(to,cap,edge[to].size()));
    ///反向边就是最后加入的那个
    edge[to].push_back(Node(from,0,edge[from].size()-1));
    ///因为上一语size句刚刚+一,故要-1
}

int level[MAXN];  //层数

bool BFS(int s, int t)
{
    queue<int>q;
    memset(level, -1, sizeof(level));
    level[s] = 0;
    q.push(s);

    while(!q.empty())
    {//cout<<"1"<<endl;
        int pre = q.front(); q.pop();

        for(int i=0; i<edge[pre].size(); ++i)////////////////////
        {
          Node &tmp1 = edge[pre][i];

         if(tmp1.cap > 0 && level[tmp1.to] == -1)
            {
                level[tmp1.to] = level[pre] + 1;
                q.push(tmp1.to);
            }

        }

    }
    return level[t] != -1;
}

int DFS(int u, int t, int flow)
{//cout<<"2"<<endl;
    if(u == t) return flow;

    int tf = 0; ///
    for(int i=0; i<edge[u].size(); ++i)////////////////////
    {
      Node &tmp2 = edge[u][i];
      if(tmp2.cap>0 && flow>tf &&level[tmp2.to]==level[u]+1)
      {
           int f = DFS(tmp2.to,t,min(flow-tf, tmp2.cap));
           if(f>0)
           {
               tmp2.cap -= f;
               edge[tmp2.to][tmp2.rev].cap += f;
               tf += f;    
           }

      }
    }
    if(tf==0) level[u] = -1;
    return tf;
}


int DINIC(int s, int t)
{
    int ans = 0;
    while(BFS(s,t))
    {
        int  tp;
        while(tp = DFS(s,t,INF))
            ans += tp;
    }
    return ans;
}






int main()
{

    int n,np,nc,m;
   // while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)
   while(cin>>n>>np>>nc>>m)
    {
        for(int i=0;i<MAXN; ++i)
            edge[i].clear();

        char t1,t2,t3;
        int u, v, cap;
        for(int i=1; i<=m; ++i)
        {
            //scanf("%c%d%c%d%c%d",&t1,&u,&t2,&v,&t3,&cap);
            cin>>t1>>u>>t2>>v>>t3>>cap;
            if(u==v) continue;
            add_node(u,v,cap);
        }

        int s = n+1;
        int t = n+2;
        for(int i=1; i<=np; ++i)
        {
            //scanf("%c%d%c%d",&t1,&v,&t2,&cap);
            cin>>t1>>v>>t2>>cap;
            add_node(s,v,cap);
        }

        for(int i=1; i<=nc; ++i)
        {
            //scanf("(%d)%d",u,cap);
            //scanf("%c%d%c%d",&t1,&u,&t2,&cap);
            cin>>t1>>u>>t2>>cap;
            add_node(u,t,cap);
        }


        //printf("%d\n", DINIC(s, t));
        cout<<DINIC(s,t)<<endl;
    }
    return 0;
}
### 关于网络流算法的模板题目 网络流是一种经典的图论算法,主要用于解决最大流、最小割等问题。以下是几个常见的网络流算法模板及其对应的经典题目。 #### 1. 最大流问题 最大流问题是网络流中最基本的问题之一,通常可以通过 **Edmonds-Karp 算法** 或者 **Dinic 算法** 解决。下面是一个简单的 Edmonds-Karp 算法实现: ```python from collections import deque class Edge: def __init__(self, v, flow, rev): self.v = v self.flow = flow self.rev = rev def add_edge(u, v, capacity, graph): edge_u_to_v = Edge(v, capacity, len(graph[v])) edge_v_to_u = Edge(u, 0, len(graph[u])) graph[u].append(edge_u_to_v) graph[v].append(edge_v_to_u) def bfs(s, t, parent, graph): visited = [False] * len(graph) queue = deque([s]) visited[s] = True while queue: u = queue.popleft() for idx, edge in enumerate(graph[u]): if not visited[edge.v] and edge.flow > 0: queue.append(edge.v) visited[edge.v] = True parent[edge.v] = u if edge.v == t: return True return False def edmonds_karp(n, s, t, graph): parent = [-1] * n max_flow = 0 while bfs(s, t, parent, graph): path_flow = float('Inf') v = t while v != s: u = parent[v] path_flow = min(path_flow, graph[u][next(i for i, e in enumerate(graph[u]) if e.v == v)].flow) v = u v = t while v != s: u = parent[v] index = next(i for i, e in enumerate(graph[u]) if e.v == v) graph[u][index].flow -= path_flow graph[v][graph[u][index].rev].flow += path_flow v = u max_flow += path_flow return max_flow ``` 上述代码实现了基于 BFS 的 Edmonds-Karp 算法来求解最大流问题[^4]。 --- #### 2. 最小费用最大流问题 如果需要考虑每条边的成本,则可以使用 **SPFA** 或 **Bellman-Ford** 结合最短路径的思想来计算最小费用的最大流。以下是一个 SPFA 实现的例子: ```python import heapq INF = int(1e9) def spfa_min_cost_max_flow(n, edges, start, end): adj_list = [[] for _ in range(n)] residual_graph = [[None]*n for _ in range(n)] for u, v, cap, cost in edges: adj_list[u].append((v, cap, cost)) adj_list[v].append((u, 0, -cost)) dist = [INF] * n potential = [0] * n prev_node = [-1] * n prev_edge = [-1] * n total_flow = 0 total_cost = 0 while True: pq = [] dist[start] = 0 heapq.heappush(pq, (dist[start], start)) while pq: d, node = heapq.heappop(pq) if d > dist[node]: continue for idx, (neighbor, cap, cost) in enumerate(adj_list[node]): if cap > 0 and dist[neighbor] > dist[node] + cost + potential[node] - potential[neighbor]: dist[neighbor] = dist[node] + cost + potential[node] - potential[neighbor] prev_node[neighbor] = node prev_edge[neighbor] = idx heapq.heappush(pq, (dist[neighbor], neighbor)) if dist[end] == INF: break for i in range(n): potential[i] += dist[i] flow = INF cur = end while cur != start: previous = prev_node[cur] edge_index = prev_edge[cur] flow = min(flow, adj_list[previous][edge_index][1]) cur = previous total_flow += flow total_cost += flow * potential[end] cur = end while cur != start: previous = prev_node[cur] edge_index = prev_edge[cur] adj_list[previous][edge_index] = ( adj_list[previous][edge_index][0], adj_list[previous][edge_index][1] - flow, adj_list[previous][edge_index][2] ) back_edge_index = None for j, (back_neighbor, _, _) in enumerate(adj_list[cur]): if back_neighbor == previous: back_edge_index = j break adj_list[cur][back_edge_index] = ( adj_list[cur][back_edge_index][0], adj_list[cur][back_edge_index][1] + flow, adj_list[cur][back_edge_index][2] ) cur = previous return total_flow, total_cost ``` 该代码通过调整势能函数优化了 SPFA,在处理负权边时更加高效[^5]。 --- #### 3. 经典模板题推荐 以下是几道经典的网络流算法模板题,适合初学者练习: 1. **POJ 1273 Drainage Ditches**: 这是一道典型的 Edmonds-Karp 算法入门题。 2. **HDU 3549 Flow Problem**: 需要使用 Dinic 算法提高效率。 3. **Codeforces Round #XXX Div.2 C**: 涉及到最小费用最大流的应用场景。 4. **LeetCode 787 Cheapest Flights Within K Stops**: 虽然不是纯网络流问题,但可以用类似思路建模。 ---
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