网络流最大流——Dinic算法

Posted 2021-03-03 ignorance

Dinic 时间复杂度最坏 O(n*n*m) 平均O（n）。

算法思路：用BFS构建层次图，如果汇点不在层次图中则结束算法返回最大流，否则在层次图中DFS找到图中所有增广路，增广结束后重新建立层次图。

 

优化：多路增广优化：将节点的所有增广路的到的流量记录下来直接返回。

　　　炸点优化：若改点无流量直接弃掉。

 

code：

#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <functional>
#include <map>
#include <set>
#include <stack>
#define FT(a, b) memset(a, b, sizeof(a))
#define FAT(a) memset(a, 0, sizeof(a))
using namespace std;
typedef long long ll;
const int M = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n, m, s, t;
int h[M], e[M], ne[M], w[M], idx;
int deepth[M];
void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
bool bfs()// bfs求深度
{
    FT(deepth, -1);
    queue<int> q;
    q.push(s);
    deepth[s] = 0;
    while (!q.empty())
    {
        int a = q.front();
        q.pop();
        for (int i = h[a]; ~i; i = ne[i])
        {
            int j = e[i];
            if (w[i] && deepth[j] == -1)//只有未被访问过且还有流量
            {
                deepth[j] = deepth[a] + 1;
                q.push(j);
            }
        }
    }
    return deepth[t] != -1;
}
int dfs(int now, int flow)
{
    if (now == t)
        return flow;
    int nowflow = 0; //多路增广优化
    for (int i = h[now]; ~i; i = ne[i])
    {
        int j = e[i];
        if (w[i] && deepth[j] == deepth[now] + 1)// 必须有流量且比上一个点深
        {
            int k = dfs(j, min(flow, w[i]));
            w[i] -= k;
            w[i ^ 1] += k;
            flow -= k;
            nowflow += k;
        }
    }
    if (!nowflow)
        deepth[now] = -2; //炸点优化
    return nowflow;
}
int dinic()
{
    int ans = 0;
    while (bfs())
        ans += dfs(s, INF);
    return ans;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("D://code//c++//in.txt", "r", stdin);
#endif
    scanf("%d%d%d%d", &n, &m, &s, &t);
    FT(h, -1);
    idx = 0;
    for (int i = 0; i < m; i++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c), add(b, a, 0);
    }
    printf("%d
", dinic());

    return 0;
}