Dinic算法求解最大流问题

题目：luogu 3376

分析：用Dinic算法求解最大流问题，复杂度为$O(mn^{2})$. 核心有两部分：利用BFS构造level graph, 然后用DFS找增广路径。代码如下，难点在于函数DFS(start, flow), 返回从结点start出发到$t$的最大流，并且满足流值不超过flow.

#include<iostream>
#include<queue>
#include<algorithm>
#include<string.h>
using namespace std;
//n<=200, m<=5000, w<=2^31-1 (int)
int utoe[205];//结点u对应的边e的编号
int level[205];//每个结点的level number
long long INF = 429496729400;
int s, t;
long long maxflow;
struct
{
    long long cap;//边的容量
    int toVer;//边(u,v)对应的v
    int nex;//边(u,v1)上一条共起点的边(u,v2)的编号
} edge[10005];
bool constructLG()//构造level graph
{
    queue<int>q;
    memset(level, -1, sizeof(level));//每个结点的level初始化为-1
    int e_idx;//边的编号
    int top_ele;//队首元素(结点编号)
    int v_temp;
    q.push(s);
    level[s] = 0;
    while (!q.empty())
    {
        top_ele = q.front();
        q.pop();
        e_idx=utoe[top_ele];
        while (e_idx >= 0)
        {
            v_temp = edge[e_idx].toVer;
            if (edge[e_idx].cap > 0 && level[v_temp] == -1)
            {
                q.push(v_temp);
                level[v_temp] = level[top_ele] + 1;
                if (v_temp == t)
                    return true;
            }
            e_idx = edge[e_idx].nex;
        }
    }
    return false;
}
long long DFS(int start, long long flow)//flow是为结点start分配的flow value
{
    long long res = 0;
    long long subflow;
    if (start == t)
        return flow;
    int e_idx=utoe[start];
    int v_temp;
    while (e_idx>=0 && flow>0)//flow如果为零，没必要再搜索下去
    {
        v_temp = edge[e_idx].toVer;
        if (level[v_temp] == level[start] + 1 && edge[e_idx].cap > 0)
        {
            subflow=DFS(v_temp,min(flow,edge[e_idx].cap));
            if (subflow == 0)//剪枝
                level[v_temp] = -1;
            edge[e_idx].cap -= subflow;
            edge[e_idx ^ 1].cap += subflow;
            res += subflow;
            flow -= subflow;
        }
        e_idx = edge[e_idx].nex;
    }
    return res;
}
void Dinic()
{
    while (constructLG())
        maxflow+=DFS(s, INF);
}
int main()
{
    int n, m;
    int u, v, c;
    int id_e;
    int i;
    while (scanf("%d%d%d%d", &n, &m, &s, &t) == 4)
    {
        id_e = -1;
        memset(utoe, -1, sizeof(utoe));
        for (i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &u, &v, &c);
            edge[++id_e].cap = c;
            edge[id_e].toVer = v;
            edge[id_e].nex = utoe[u];
            utoe[u] = id_e;

            edge[++id_e].cap = 0;
            edge[id_e].toVer = u;
            edge[id_e].nex = utoe[v];
            utoe[v] = id_e;
        }
        maxflow = 0;
        Dinic();
        printf("%lld\n", maxflow);
    }
    return 0;
}