最小费用最大流（SPFA版本）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;

#define LL long long
#define pb push_back
#define fi first
#define se second


//时间复杂度: O(V * E^2 * log(V))
//SPFA中不会出现负环，如果出现环，则说明这些边还有流量，
//但是每次流过之后这些边的流量都会减去，加上反向边


const int N = 5e3 + 10;
const int M = 5e4 + 10;
const int INF = 1e9;

struct edge{
    int to, nxt, cap ,flow, cost;
}e[M << 2];
int head[N], vis[N], dis[N], pre[N];
queue<int > que;
int n, m, s, t, tot;

inline void add(int u, int v, int cap, int w){
    e[tot].to = v; e[tot].cap = cap; e[tot].cost = w;
    e[tot].flow = 0; e[tot].nxt = head[u]; head[u] = tot++;
    e[tot].to = u; e[tot].cap = 0; e[tot].cost = -w;
    e[tot].flow = 0; e[tot].nxt = head[v]; head[v] = tot++;
}

void init(){
    for(int i = 0; i <= n; ++i){
        dis[i] = INF;
        vis[i] = 0;
        pre[i] = -1;
    }
    while(!que.empty()) que.pop();
}

bool SPFA(){
    init();
    dis[s] = 0;
    vis[s] = 1;
    que.push(s);

    int now, to, cap, flow, cost;
    while(!que.empty()){
        now = que.front();
        que.pop();
        vis[now] = 0;

        for(int o = head[now]; ~o; o = e[o].nxt){
            to = e[o].to;
            cap =  e[o].cap;
            flow = e[o].flow;
            cost = e[o].cost;

            if(cap - flow > 0 && dis[to] > dis[now] + cost){
                dis[to] = dis[now] + cost;

                //记录来源
                pre[to] = o;

                if(!vis[to]){
                    vis[to] = 1;
                    que.push(to);
                }
            }
        }    
    }

    if(pre[t] == -1) return false;
    else return true;
}

void mcmf(){
    
    int mf = 0, mc = 0;    
    while(SPFA()){
        int Min = INF;

        //找到最窄的水管
        for(int o = pre[t]; ~o; o = pre[e[o ^ 1].to]){
            Min = min(Min, e[o].cap - e[o].flow);
        }

        //更新流量
        for(int o = pre[t]; ~o; o = pre[e[o ^ 1].to]){
            e[o].flow += Min;
            e[o ^ 1].flow -= Min;
        }
        mf += Min;
        mc += Min * dis[t];
    }

    printf("%d %d
", mf, mc);
}

void solve(){
    scanf("%d%d%d%d", &n, &m, &s, &t);
    for(int i = 0; i <= n; ++i) head[i] = -1; tot = 0;
    
    int u, v, w, f;
    for(int i = 0; i < m; ++i){
        scanf("%d%d%d%d", &u, &v, &w, &f);
        add(u, v, w, f);
    }
    mcmf();
}

int main(){

    solve();

    return 0;
}