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;
}

  

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