MCMF(最小费用最大流):由于要使用反向边。定义一个h[i]:表示从汇点到i的最短距离。对于一条边e(v,u),e.cost = e.cost + h[v] - h[u],这样可以保证图中没有负权边的存在,这样就可以使用dij来求最短路。
每次求一条最短路,然后通过这条最短路更新最大流,直到找不出最短路为止。
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define PI acos(-1)
#define INF 0x3f3f3f3f
#define NUM 50010
#define debug true
#define lowbit(x) ((-x)&x)
#define ffor(i,d,u) for(int i=d;i<=u;++i)
#define _ffor(i,u,d) for(int i=u;i>=d;--i)
#define mst(array,Num) memset(array,Num,sizeof(array))
const int p = 1e9+7;
int n,m,ednum;
int head[NUM/10],h[NUM/10]={};
int pre[NUM/10],dist[NUM/10];
struct edge
{
int next,to,w,c;
}e[NUM<<1];
struct Vertex
{
int id,dis;
bool operator>(const Vertex &x)const
{
return dis > x.dis ;
}
};
template <typename T>
inline void read(T &x){
char ch = getchar();x = 0;
for (; ch < '0' || ch > '9'; ch = getchar());
for (; ch >='0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
}
template <typename T>
inline void write(T x)
{
int len=0;char c[21];
if(x<0)putchar('-'),x*=(-1);
do{++len;c[len]=(x%10)+'0';}while(x/=10);
_ffor(i,len,1)putchar(c[i]);
}
inline bool Dij(const int &s,const int &t)
{
Vertex x,y;
priority_queue < Vertex , vector < Vertex > , greater < Vertex > > q;
mst(dist,INF);
x.id = s,dist[s] = x.dis = 0;
pre[s] = -1;
q.push(x);
while(!q.empty())
{
x = q.top(),q.pop();
if(x.dis > dist[x.id])continue;
for(int i = head[x.id] ; i != -1 ; i = e[i].next)
{
y.id = e[i].to;
if(e[i].w > 0 && dist[y.id] > dist[x.id]+e[i].c+h[x.id]-h[y.id])
{
y.dis = dist[y.id] = dist[x.id]+e[i].c+h[x.id]-h[y.id];
pre[y.id] = i;
q.push(y);
}
}
}
if(dist[t] == INF)return false;
return true;
}
inline void MinCost_MaxFlow(const int &s,const int &t,int &cost,int &maxflow)
{
int d=INF;
mst(h,0);
while(d > 0&&Dij(s,t))
{
ffor(i,1,n) h[i] += dist[i];
d = INF;
for(int i = pre[t] ; i != -1 ; i = pre[e[i^1].to])
d = min(d , e[i].w);
maxflow += d , cost += h[t] * d;
for(int i = pre[t] ; i != -1 ; i = pre[e[i^1].to])
{
e[i].w -= d;
e[i^1].w += d;
}
}
}
inline void AC()
{
int x,y,wei,s,t;
int cost,maxflow=0;
read(n),read(m),read(s),read(t);
maxflow = 0,ednum = -1,mst(head,-1);
ffor(i,1,m)
{
read(x),read(y),read(wei),read(cost);
e[++ednum].to = y,e[ednum].w = wei,e[ednum].next = head[x],head[x] = ednum,e[ednum].c = cost;
e[++ednum].to = x,e[ednum].w = 0,e[ednum].next = head[y],head[y] = ednum,e[ednum].c = -cost;
}
cost = 0;
MinCost_MaxFlow(s,t,cost,maxflow);
write(maxflow),putchar(' '),write(cost),puts("");
}
int main()
{
AC();
return 0;
}