上下界流

3.上下界流

无源汇上下界可行流

给有向图 G， 每条边都有一个流量上界和流量下界。若存在可行流，则输出每条边的流量，若不存在输出“NO“

思路：

对于每条边，先流下界，统计每个点的流量，所有点的流量和一定是0，建立一个超级源点连接所有流量为正的点，超级汇点连接流量为负的点，跑最大流，看能不能跑满流。

/*
 * @Author: zhl
 * @Date: 2020-10-20 11:09:59
 */

int val[N];
int minF[N];
signed main() {
	scanf("%d%d", &n, &m);
	memset(head, -1, sizeof(int) * (n + 10));
	for (int i = 1; i <= m; i++) {
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		addEdge(a, b, d - c);
		val[a] -= c;
		val[b] += c;
		minF[i - 1] = c;
	}
	int sum = 0;
	for (int i = 1; i <= n; i++) {
		if (val[i] > 0)addEdge(0, i, val[i]), sum += val[i];
		if (val[i] < 0)addEdge(i, n + 1, -val[i]);
	}
	s = 0, t = n + 1;
	int maxflow = Dinic();

	if (maxflow == sum) {
		printf("YES
");
		for (int i = 0; i < m; i++) {
			printf("%d
", E[(2*i)^1].flow + minF[i]);
		}
	}
	else {
		printf("NO
");
	}
}

有源汇上下界最大流

还是像无源汇上下界那样。

addEdge(t,s,inf) ，这么一加，然后跑一遍超级源点到超级汇点的可行流

加的这条边的反向边，就是一个s 到 t 的基础流

然后再删掉这边，在此时的残留网络中跑一遍 s 到 t 的最大流

两个答案加起来就是 有源汇上下界最大流

/*
 * @Author: zhl
 * @Date: 2020-10-20 11:09:59
 */
#include<bits/stdc++.h>
 //#define int long long
using namespace std;

const int N = 1e4 + 10, M = 1e5 + 10, inf = 1e9;
int n, m, s, t, tot, head[N];
int ans, dis[N], cur[N];

struct Edge {
	int to, next, flow;
}E[M << 1];

void addEdge(int from, int to, int w) {
	E[tot] = Edge{ to,head[from],w };
	head[from] = tot++;
	E[tot] = Edge{ from,head[to],0 };
	head[to] = tot++;
}

int bfs() {
	for (int i = 0; i <= n + 1; i++) dis[i] = -1;
	queue<int>Q;
	Q.push(s);
	dis[s] = 0;
	cur[s] = head[s];

	while (!Q.empty()) {
		int u = Q.front();
		Q.pop();
		for (int i = head[u]; ~i; i = E[i].next) {
			int v = E[i].to;
			if (E[i].flow && dis[v] == -1) {
				Q.push(v);
				dis[v] = dis[u] + 1;
				cur[v] = head[v];
				if (v == t)return 1; //分层成功
			}
		}
	}
	return 0;
}

int dfs(int x, int sum) {
	if (x == t)return sum;
	int k, res = 0;
	for (int i = cur[x]; ~i && res < sum; i = E[i].next) {
		cur[x] = i;
		int v = E[i].to;
		if (E[i].flow > 0 && (dis[v] == dis[x] + 1)) {
			k = dfs(v, min(sum, E[i].flow));
			if (k == 0) dis[v] = -1; //不可用
			E[i].flow -= k; E[i ^ 1].flow += k;
			res += k; sum -= k;
		}
	}
	return res;
}

int Dinic() {
	int ans = 0;
	while (bfs()) {
		ans += dfs(s, inf);
	}
	return ans;
}

int val[N];
int minF[N];
signed main() {
	int S, T;
	scanf("%d%d%d%d", &n, &m, &S, &T);
	memset(head, -1, sizeof(int) * (n + 10));
	for (int i = 1; i <= m; i++) {
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		addEdge(a, b, d - c);
		val[a] -= c;
		val[b] += c;
		minF[i - 1] = c;
	}
	int sum = 0;
	for (int i = 1; i <= n; i++) {
		if (val[i] > 0)addEdge(0, i, val[i]), sum += val[i];
		if (val[i] < 0)addEdge(i, n + 1, -val[i]);
	}
	s = 0, t = n + 1;
	addEdge(T, S, inf);
	int maxflow = Dinic();

	if (maxflow == sum) {
		int base = E[tot - 1].flow;
		E[tot - 1].flow = E[tot - 2].flow = 0;
		s = S; t = T;
		printf("%d
", base + Dinic());
	}
	else {
		printf("No Solution
");
	}
}

有源汇上下界最小流

跟上面类似，跑 t 到 s 的最大流，减去就可以了

/*
 * @Author: zhl
 * @Date: 2020-10-20 11:09:59
 */
#include<bits/stdc++.h>
 //#define int long long
using namespace std;

const int N = 2e6 + 10, M = 2e6 + 10, inf = 1e9;
int n, m, s, t, tot, head[N];
int ans, dis[N], cur[N];

struct Edge {
	int to, next, flow;
}E[M << 1];

void addEdge(int from, int to, int w) {
	E[tot] = Edge{ to,head[from],w };
	head[from] = tot++;
	E[tot] = Edge{ from,head[to],0 };
	head[to] = tot++;
}

int bfs() {
	for (int i = 0; i <= n + 1; i++) dis[i] = -1;
	queue<int>Q;
	Q.push(s);
	dis[s] = 0;
	cur[s] = head[s];

	while (!Q.empty()) {
		int u = Q.front();
		Q.pop();
		for (int i = head[u]; ~i; i = E[i].next) {
			int v = E[i].to;
			if (E[i].flow && dis[v] == -1) {
				Q.push(v);
				dis[v] = dis[u] + 1;
				cur[v] = head[v];
				if (v == t)return 1; //分层成功
			}
		}
	}
	return 0;
}

int dfs(int x, int sum) {
	if (x == t)return sum;
	int k, res = 0;
	for (int i = cur[x]; ~i && res < sum; i = E[i].next) {
		cur[x] = i;
		int v = E[i].to;
		if (E[i].flow > 0 && (dis[v] == dis[x] + 1)) {
			k = dfs(v, min(sum, E[i].flow));
			if (k == 0) dis[v] = -1; //不可用
			E[i].flow -= k; E[i ^ 1].flow += k;
			res += k; sum -= k;
		}
	}
	return res;
}

int Dinic() {
	int ans = 0;
	while (bfs()) {
		ans += dfs(s, inf);
	}
	return ans;
}

int val[N];
int minF[N];
signed main() {
	int S, T;
	scanf("%d%d%d%d", &n, &m, &S, &T);
	memset(head, -1, sizeof(int) * (n + 10));
	for (int i = 1; i <= m; i++) {
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		addEdge(a, b, d - c);
		val[a] -= c;
		val[b] += c;
		minF[i - 1] = c;
	}
	int sum = 0;
	for (int i = 1; i <= n; i++) {
		if (val[i] > 0)addEdge(0, i, val[i]), sum += val[i];
		if (val[i] < 0)addEdge(i, n + 1, -val[i]);
	}
	s = 0, t = n + 1;
	addEdge(T, S, inf);
	int maxflow = Dinic();

	if (maxflow == sum) {
		int base = E[tot - 1].flow;
		E[tot - 1].flow = E[tot - 2].flow = 0;
		s = T; t = S;//只有这里改动了
		printf("%d
", base - Dinic());
	}
	else {
		printf("No Solution
");
	}
}