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  • poj1273--Drainage Ditches(最大流Edmond-Karp算法 邻接表实现)

    最大流模板题

    大部分Edmond-Karp算法代码都是邻接矩阵实现,试着改成了邻接表。

    #include <iostream>
    #include <cstdio>
    #include <queue>
    #include <cstring>
    
    using namespace std;
    
    // 裸最大流
    const int N = 2005;
    const int M = 2005;
    const int INF = 0x7fffffff;
    
    // 邻接表
    struct Edge {
        int from, to, next, cost;
    } edge[M];
    int head[N];
    int cnt_edge;
    void add_edge(int u, int v, int c)
    {
        edge[cnt_edge].to = v;
        edge[cnt_edge].from = u;
        edge[cnt_edge].cost = c;
        edge[cnt_edge].next = head[u];
        head[u] = cnt_edge++;
    }
    
    int pre[N];
    int flow[N];
    queue<int> q;
    int bfs(int src, int des)
    {
        memset(pre, -1, sizeof pre);
        while (!q.empty()) q.pop();
        q.push(src);
        flow[src] = INF;
        while (!q.empty())
        {
            int u = q.front();
            q.pop();
            if (u == des) break;
            for (int i = head[u]; i != -1; i = edge[i].next)
            {
                int v = edge[i].to;
                int cost = edge[i].cost;
                if (pre[v] == -1 && cost > 0)
                {
                    flow[v] = min(flow[u], cost);
                    pre[v] = i; // 记录的是边
                    q.push(v);
                }
            }
        }
        if (pre[des] == -1) return -1;
        return flow[des];
    }
    
    int maxFlow(int src, int des)
    {
        int ans = 0;
        int in;
        while ((in = bfs(src, des)) != -1)
        {
            int k = des;
            while (k != src)
            {
                int last = pre[k];
                edge[last].cost -= in;
                edge[last ^ 1].cost += in;
                k = edge[last].from;
            }
            ans += in;
        }
        return ans;
    }
    
    int main()
    {
        int n, m;
        while (~scanf("%d%d", &m, &n))
        {
            int a, b, c;
            memset(head, -1, sizeof head);
            while (m--)
            {
                scanf("%d%d%d", &a, &b, &c);
                add_edge(a, b, c);
                add_edge(b, a, 0);
            }
            printf("%d
    ", maxFlow(1, n));
        }
    }
    

      

    另附邻接矩阵版

    #include <iostream>
    #include <cstdio>
    #include <queue>
    #include <cstring>
    #define pk puts("kk");
    
    using namespace std;
    
    // 裸最大流
    const int N = 205;
    const int INF = 0x7fffffff;
    
    int cap[N][N];
    int flow[N];
    int pre[N];
    queue<int> q;
    
    int n, m;
    
    int bfs(int src, int des)
    {
        while (!q.empty()) q.pop();
        memset(pre, -1, sizeof pre);
        q.push(src);
        flow[src] = INF;
        while (!q.empty())
        {
            int idx = q.front();
            if (idx == des) break;
            q.pop();
            for (int i = 1; i <= n; ++i)
            {
                if (pre[i] == -1 && cap[idx][i] > 0)
                {
                    flow[i] = min(flow[idx], cap[idx][i]);
                    pre[i] = idx;
                    q.push(i);
                }
            }
        }
        if (pre[des] == -1) return -1;
        return flow[des];
    }
    
    int maxFlow(int src, int des)
    {
        int ans = 0;
        int in;
        while ((in = bfs(src, des)) != -1)
        {
            int k = des;
            while (k != src)
            {
                int last = pre[k];
                cap[last][k] -= in;
                cap[k][last] += in;
                k = last;
            }
            ans += in;
        }
        return ans;
    }
    
    int main()
    {
        while (~scanf("%d%d", &m, &n))
        {
            int a, b, c;
            memset(cap, 0, sizeof cap);
            for (int i = 0; i < m; ++i)
            {
                scanf("%d%d%d", &a, &b, &c);
                cap[a][b] += c;
            }
            printf("%d
    ", maxFlow(1, n));
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/wenruo/p/5013136.html
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