#include<algorithm>
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
ll gcd(ll a, ll b) {
return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B) {
(A<B)&&(A=B);//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B) {
(A>B)&&(A=B);//if(B<A)A=B;
}
template <class T>
inline bool scan_d(T &ret) {
char c;
int sgn;
if(c=getchar(),c==EOF) return 0; //EOF
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
inline void outt(int x) {
if(x>9) outt(x/10);
putchar(x%10+'0');
}
const int MX = 1005;
const int inf = 0x3f3f3f3f;
const int MXE = MX * MX * 4;
struct MinCost_MaxFlow {
struct Edge {
int v, w, nxt;
int cost;
} E[MXE];
int head[MX], tot, level[MX], pre[MX], d[MX];
bool vis[MX];
void init() {
memset(head, -1, sizeof(head));
tot = 0;
}
void add(int u, int v, int w, int cost) {
E[tot].v = v;
E[tot].w = w;
E[tot].cost = cost;
E[tot].nxt = head[u];
head[u] = tot++;
E[tot].v = u;
E[tot].w = 0;
E[tot].cost = -cost;
E[tot].nxt = head[v];
head[v] = tot++;
}
bool spfa(int s, int t) {
memset(vis, 0, sizeof(vis));
memset(d, 0x3f, sizeof(d));
memset(pre, -1, sizeof(pre));
queue<int>q;
q.push(s);
d[s] = 0;
vis[s] = 1;
while (!q.empty()) {
int u = q.front();
q.pop();
vis[u] = 0;
for (int i = head[u]; ~i; i = E[i].nxt) {
int w = E[i].w, v = E[i].v, cost = E[i].cost;
if (w > 0 && d[v] > d[u] + cost) {
d[v] = d[u] + cost;
pre[v] = i;
if (!vis[v]) {
q.push(v);
vis[v] = 1;
}
}
}
}
//如果是最小费用可行流则要这一句(要求费用最小,不要求流量最大)
//if (d[t] > 0) return false;
return pre[t] != -1;
}
int solve(int s, int t, int &cost) {
int flow = 0;
cost = 0;
while (spfa(s, t)) {
int minFlow = inf;
for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v])
minFlow = min(minFlow, E[i].w);
for (int i = pre[t]; ~i; i = pre[E[i ^ 1].v]) {
cost += minFlow * E[i].cost;
E[i].w -= minFlow;
E[i ^ 1].w += minFlow;
}
flow += minFlow;
}
return flow;
}
} F;
int n, m, k;
int u[10010],v[10010],w[10010],c[10010];
int main() {
scanf("%d%d%d", &n, &m, &k);
F.init();
rep(i,1,m) {
scanf("%d%d%d%d", &u[i], &v[i], &w[i], &c[i]);
F.add(u[i],v[i],w[i],0);
}
int cost=0;
int ans = F.solve(1,n,cost);
printf("%d ", ans);
rep(i,0,F.tot-1) {
if(F.tot%2) {
F.E[i].w = 0;
}
}
rep(i,1,m) {
F.add(u[i],v[i],k,c[i]);
}
F.add(0,1,k,0);
cost = 0;
F.solve(0,n,cost);
printf("%d\n", cost);
return 0;
}
撰写