最小费用最大流
Description
给出一个\(N\)个点\(M\)条边的有向图,每条边上有一个容量限制\(cap\)和单位流量的花费\(cost\)。给出源点\(s\)和汇点\(t\),求从源点\(s\)到汇点\(t\)的花费最小的最大流。输出最小花费和最大流的值。
Solution
如果没有最小花费的限制,只需要不断在残余网络上找增广路即可,如果有最小花费限制,就把每次简单找增广路改为,在以花费为权值的图上,找从\(s\)到\(t\)的最短路,每次沿最短路增广。这样就保证了每次增广得到的都是在当前流量下花费最小的流。
Code
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 1e3 + 10;
const int M = 2e3 + 10;
struct Edge
{
int from, to, next, cap, cost;
Edge() {}
Edge(int from, int to, int next, int cap, int cost) :
from(from), to(to), next(next), cap(cap), cost(cost) {}
} edge[M];
int head[N], tot;
void add(int from, int to, int cap, int cost)
{
edge[tot] = Edge(from, to, head[from], cap, cost);
head[from] = tot++;
edge[tot] = Edge(to, from, head[to], 0, -cost);
head[to] = tot++;
}
void init()
{
memset(head, -1, sizeof(head));
tot = 0;
}
int dis[N], pre[N];
bool vis[N];
queue<int> q;
bool spfa(int s, int t)
{
while (!q.empty()) q.pop();
memset(dis, 0x3f, sizeof(dis));
memset(vis, false, sizeof(vis));
memset(pre, -1, sizeof(pre));
q.push(s); vis[s] = true; dis[s] = 0;
while (!q.empty())
{
int u = q.front(); q.pop(); vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
if (edge[i].cap && dis[u] + edge[i].cost < dis[edge[i].to])
{
int v = edge[i].to;
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v]) q.push(v), vis[v] = true;
}
}
return dis[t] < INF;
}
int mcmf(int s, int t, int& maxflow)
{
int mincost = 0;
maxflow = 0;
while (spfa(s, t))
{
int flow = INF;
for (int i = pre[t]; i != -1; i = pre[edge[i].from])
flow = min(flow, edge[i].cap);
for (int i = pre[t]; i != -1; i = pre[edge[i].from])
{
edge[i].cap -= flow;
edge[i ^ 1].cap += flow;
mincost += edge[i].cost * flow;
}
maxflow += flow;
}
return mincost;
}
int main()
{
int n, m;
scanf("%d%d", &n, &m);
init();
while (m--)
{
int u, v, cap, cost;
scanf("%d%d%d%d", &u, &v, &cap, &cost);
add(u, v, cap, cost);
}
int s, t, maxflow;
scanf("%d%d", &s, &t);
int mincost = mcmf(s, t, maxflow);
printf("%d %d\n", mincost, maxflow);
return 0;
}
Input
第一行给出两个整数n和m,表示结点数和边数。
接下来m行,每行给出4个整数u,v,w,c,表示从u到v有一条容量为w,单位流量花费为c的边。
最后一行给出两个整数s和t,表示源点和汇点。
Output
输出两个整数,分别表示最小花费和最大流。
Sample Input
5 5
1 2 2 2
1 3 3 2
2 4 3 1
3 4 2 3
4 5 2 1
1 5
Sample Output
8 2