【模板】(最小费用)最大流

网络最大流

P3376 【模板】网络最大流

一、(Edmonds-Karp)算法

不断在残量网络(dfs)进行增广，当不能进行增广时，即为最大流。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 1e4 + 5, M = 2e5 + 5;
int n, m, s, t, pre[M], vis[N];
int cnt = 1, head[N], to[M], nxt[M], val[M];
void add(int u, int v, int w) {
	to[++ cnt] = v; val[cnt] = w; nxt[cnt] = head[u]; head[u] = cnt;
}
bool bfs() {
	queue <int> q; q.push(s);
	memset(vis, 0, sizeof(vis));
	memset(pre, 0, sizeof(pre));
	vis[s] = 1;
	while(! q.empty()) {
		int x = q.front(); q.pop();
		if(x == t) return true;
		for(int i = head[x]; i ; i = nxt[i]) {
			if(vis[to[i]] == 1 || val[i] == 0) continue;
			pre[to[i]] = i;
			q.push(to[i]), vis[to[i]] = 1; 
		}
	}
	return false;
}
int EK() {
	int ans = 0;
	while(bfs()) {
		int up = 2e9;
		for(int p = t; p != s; p = to[pre[p] ^ 1]) up = min(up, val[pre[p]]);
		for(int p = t; p != s; p = to[pre[p] ^ 1]) {
			val[pre[p]] -= up;
			val[pre[p] ^ 1] += up;
		}
		ans += up;
	}
	return ans;	
}
int main() {
	
	cin >> n >> m >> s >> t;
	for(int i = 1, u, v, w; i <= m; i ++) {
		cin >> u >> v >> w;
		add(u, v, w); add(v, u, 0);
	}
	cout << EK() << endl;
	return 0;
}

常数巨大，我会告诉你我的(EK)跑了544ms吗？

二、(Dinc)算法

在残量网络上(bfs)构建分层图，再从分层图上(dfs)不断增广。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 1e4 + 5, M = 2e5 + 5, INF = 2e9;
int n, m, s, t, flow, dep[N];
int cnt = 1, head[N], to[M], nxt[M], val[M];
void add(int u, int v, int w) {
	to[++ cnt] = v; val[cnt] = w; nxt[cnt] = head[u]; head[u] = cnt;
}
bool bfs() {
	memset(dep, 0, sizeof(dep));
	queue <int> q; q.push(s);
	dep[s] = 1;
	while(! q.empty()) {
		int x = q.front(); q.pop();
		for(int i = head[x]; i ; i = nxt[i]) {
			if(dep[to[i]] || val[i] <= 0) continue;
			dep[to[i]] = dep[x] + 1;
			q.push(to[i]);
		}
	}
	return dep[t] > 0;
}
int dfs(int x, int flow) {
	if(x == t) return flow;
	int up = 0;
	for(int i = head[x]; i ; i = nxt[i]) {
		if(val[i] == 0 || dep[to[i]] != dep[x] + 1) continue;
		int tmp = dfs(to[i], min(flow, val[i]));
		if(tmp == 0) continue;
		val[i] -= tmp;
		val[i ^ 1] += tmp;
		up += tmp; flow -= tmp;
		if(flow == 0) return up;
	}
	return up;

}
int Dinic() {
	int ans = 0;
	while(bfs()) while(flow = dfs(s, INF)) ans += flow;
	return ans;
}
int main() {
	cin >> n >> m >> s >> t;
	for(int i = 1, u, v, w; i <= m; i ++) {
		cin >> u >> v >> w;
		add(u, v, w); add(v, u, 0);
	}
	cout << Dinic() << endl;
	return 0;
}

其实我的(Dinic)常数更大，它跑了467ms。

不过，在加入当前弧优化和快读后，它跑了(148ms).(只加当前弧优化时跑(287ms))

当前弧优化：加速跳过枚举不必要的边。

注意在每次(bfs)前对(cur)重新赋值。

核心code:

bool bfs() {
	for(int i = 1;i <= n;i ++) cur[i] = head[i];
	memset(dep, 0, sizeof(dep));
	queue <int> q; q.push(s);
	dep[s] = 1;
	while(! q.empty()) {
		int x = q.front(); q.pop();
		for(int i = head[x]; i ; i = nxt[i]) {
			if(dep[to[i]] || val[i] <= 0) continue;
			dep[to[i]] = dep[x] + 1;
			q.push(to[i]);
		}
	}
	return dep[t] > 0;
}
int dfs(int x, int flow) {
	if(x == t) return flow;
	int up = 0;
	for(int &i = cur[x]; i ; i = nxt[i]) {
		if(val[i] == 0 || dep[to[i]] != dep[x] + 1) continue;
		int tmp = dfs(to[i], min(flow, val[i]));
		if(tmp == 0) continue;
		val[i] -= tmp;
		val[i ^ 1] += tmp;
		up += tmp; flow -= tmp;
		if(flow == 0) return up;
	}
	return up;
}

最小费用最大流

P3381 【模板】最小费用最大流

最小费用流

将(bfs)改为(spfa)跑最小费用即可。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 1e4 + 5, M = 2e5 + 5;
int n, m, s, t, ans, res, pre[M], vis[N], dis[N];
int cnt = 1, head[N], to[M], nxt[M], val[M], cost[M];
void add(int u, int v, int w, int c) {
	to[++ cnt] = v; val[cnt] = w; cost[cnt] = c;
	nxt[cnt] = head[u]; head[u] = cnt;
}
bool spfa() {
	queue <int> q; q.push(s);
	memset(vis, 0, sizeof(vis));
	memset(dis, 0x3f, sizeof(dis));
	vis[s] = 1; dis[s] = 0; 
	while(! q.empty()) {
		int x = q.front(); q.pop();
		vis[x] = false;
		for(int i = head[x]; i ; i = nxt[i]) {
			if(val[i] && dis[to[i]] > dis[x] + cost[i]) {
				pre[to[i]] = i;
				dis[to[i]] = dis[x] + cost[i];
				if(vis[to[i]] == 0) q.push(to[i]), vis[to[i]] = true;
			}
		}
	}
	return dis[t] != 0x3f3f3f3f;
}
void EK() {
	while(spfa()) {
		int up = 2e9;
		for(int p = t; p != s; p = to[pre[p] ^ 1]) up = min(up, val[pre[p]]);
		for(int p = t; p != s; p = to[pre[p] ^ 1]) {
			val[pre[p]] -= up;
			val[pre[p] ^ 1] += up;
		}
		res += up * dis[t];
		ans += up;
	}	
}
int main() {
	cin >> n >> m >> s >> t;
	for(int i = 1, u, v, w, c; i <= m; i ++) {
		cin >> u >> v >> w >> c;
		add(u, v, w, c); add(v, u, 0, -c);
	}
	EK();
	cout << ans << " " << res << endl;
	return 0;
}