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题意：n个节点m条有向边，每条有向边的容量是C,且费用是a*x^2，x是流量，问从1运送k流量到n的最小费用
一般做的费用流边的费用都是固定的，而这题并不是固定的。
但是，看到了C<=5，其实就是在提示可以拆边..
假如C是3，我们就可以把一条边拆成3条边。
假如不拆，如果通过的流量是1，2，3，那么费用分别是a，4a，9a
如果拆成3条边，那么3条边的费用分别是a，3a，5a，容量都是1
这样就完美的解决了边的费用问题了~
#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cctype>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)using namespace std;
typedef long long LL;
typedef pair<int, int> PII;const int MX = 1e2 + 5;
const int MM = 2e5 + 5;
const int INF = 0x3f3f3f3f;struct Edge {int to, next, cap, flow, cost;Edge() {}Edge(int _to, int _next, int _cap, int _flow, int _cost) {to = _to; next = _next; cap = _cap; flow = _flow; cost = _cost;}
} E[MM];int Head[MX], tol;
int pre[MX]; //储存前驱顶点
int dis[MX]; //储存到源点s的距离
bool vis[MX];
int N;//节点总个数，节点编号从0~N-1void init(int n) {tol = 0;N = n + 2;memset(Head, -1, sizeof(Head));
}
void edge_add(int u, int v, int cap, int cost) {E[tol] = Edge(v, Head[u], cap, 0, cost);Head[u] = tol++;E[tol] = Edge(u, Head[v], 0, 0, -cost);Head[v] = tol++;
}
bool spfa(int s, int t) {queue<int>q;for (int i = 0; i < N; i++) {dis[i] = INF;vis[i] = false;pre[i] = -1;}dis[s] = 0;vis[s] = true;q.push(s);while (!q.empty()) {int u = q.front();q.pop();vis[u] = false;for (int i = Head[u]; i != -1; i = E[i].next) {int v = E[i].to;if (E[i].cap > E[i].flow && dis[v] > dis[u] + E[i].cost) {dis[v] = dis[u] + E[i].cost;pre[v] = i;if (!vis[v]) {vis[v] = true;q.push(v);}}}}if (pre[t] == -1) return false;else return true;
}//返回的是最大流， cost存的是最小费用
int minCostMaxflow(int s, int t, int &cost) {int flow = 0;cost = 0;while (spfa(s, t)) {int Min = INF;for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {if (Min > E[i].cap - E[i].flow)Min = E[i].cap - E[i].flow;}for (int i = pre[t]; i != -1; i = pre[E[i ^ 1].to]) {E[i].flow += Min;E[i ^ 1].flow -= Min;cost += E[i].cost * Min;}flow += Min;}return flow;
}inline int read() {char c = getchar();while(!isdigit(c)) c = getchar();int x = 0;while(isdigit(c)) {x = x * 10 + c - '0';c = getchar();}return x;
}int dist[] = {0, 1, 3, 5, 7, 9};int main() {int n, m, k; //FIN;while(~scanf("%d%d%d", &n, &m, &k)) {int s = 0, t = n + 1;init(t);for(int i = 1; i <= m; i++) {int u, v, a, c;scanf("%d%d%d%d", &u, &v, &a, &c);for(int j = 1; j <= c; j++) {edge_add(u, v, 1, a * dist[j]);}}edge_add(s, 1, k, 0);edge_add(n, t, k, 0);int ans = 0;if(minCostMaxflow(s, t, ans) != k) {printf("-1\n");} else printf("%d\n", ans);}return 0;
}