luogu3381 【模板】最小费用最大流

题面在这里

题意：

给一个有向图，给定边上的容量和费用，给定源点汇点，求最大流和最小费用。

做法：

【模板】
具体不解释了（网上有很多资料，自己百度，有空再来科普qaq）。
有一个地方就是修改边的容量的时候，我原先写的都是+1-1的修改，不过洛谷上那道模板过不去的，所以要先找到一个最小的流量d然后每次+d-d，这样才能过。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cctype>
#include<queue>
using namespace std;
typedef long long ll;

inline ll read() {
    char ch = getchar(); ll x = 0; int op = 1;
    for(; !isdigit(ch); ch = getchar()) if(ch == '-') op = -1;
    for(; isdigit(ch); ch = getchar()) x = x*10+ch-'0';
    return x*op;
}
inline void write(ll a) {
    if(a < 0) putchar('-'), a = -a;
    if(a >= 10) write(a/10); putchar('0'+a%10);
}

const int N = 50010;
const int inf = 1e9;
int n, m, s, t, cnt, ans, ans1;
int dis[N], vis[N], head[N], pre[N];
struct edge {
    int to, nxt, c, w;
    edge() {}
    edge(int u, int v, int x, int y) { to = u, nxt = v, c = x, w = y; }
}e[N<<1];

inline void addedge(int x, int y, int c, int w) { e[cnt] = edge(y, head[x], c, w); head[x] = cnt ++; }
inline bool spfa() {
    queue<int> q; q.push(s);
    for(int i = 1; i <= n; i ++) dis[i] = inf; memset(vis, 0, sizeof vis);
    dis[s] = 0; vis[s] = 1;
    while(!q.empty()) {
        int u = q.front(); q.pop(); vis[u] = 0;// printf("%d\n", u);
        for(int i = head[u]; i != -1; i = e[i].nxt) {
            int v = e[i].to;
            if(e[i].c > 0 && dis[v] > dis[u]+e[i].w) {
                dis[v] = dis[u]+e[i].w; pre[v] = i;
                if(!vis[v]) {
                    vis[v] = 1; q.push(v);
                }
            }
        }
    }
    if(dis[t] == inf) return 0;
    int d = inf;
    for(int i = t; i != s; ) {
        int t = pre[i]; d = min(d, e[t].c);
        i = e[t^1].to;
    }
    ans1 += d; ans += d*dis[t];
    for(int i = t; i != s; ) {
        int t = pre[i]; e[t].c -= d; e[t^1].c += d;
        i = e[t^1].to;
    } return 1;
}
int main() {
    //freopen("testdata.in", "r", stdin);
    n = read(), m = read(), s = read(), t = read(); memset(head, -1, sizeof head);
    for(int i = 1; i <= m; i ++) {
        int x = read(), y = read(), z = read(), w = read();
        addedge(x, y, z, w); addedge(y, x, 0, -w);
    }
    while(spfa());
    printf("%d %d\n", ans1, ans);
    return 0;
}