Luo Gu P3376-- [template] network maximum flow

Topic Portal

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Title Description

If that, given a network diagram, and the source and sink, the network obtains its maximum flow.

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Input Format

The first row contains four positive integers N, M, S, T, respectively, the number of points, there are the number of edges, number source, sink point number.

Next M lines contains three positive integers ui, vi, wi, represents the i ui article there from edge to reach VI, the right side of wi (i.e., the side of the maximum flow rate wi)

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Output Format

Line, contains a positive integer, is the maximum flow of the network.

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Entry

4 5 4 3
4 2 30
4 3 20
2 3 20
2 1 30
1 3 40

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Export

50

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Description / Tips

Constraints of time: 1000ms, 128M

Data Scale:

For 30% of the data: N <= 10, M <= 25

For 70% of the data: N <= 200, M <= 1000

To 100% of the data: N <= 10000, M <= 100000

Sample Description:

There are three paths title:

4-> 2-> 3, the line 20 can flow through the

4-> 3, the flow 20 may be prepared by

4-> 2-> 1-> 3, the flow 10 may be prepared by (edge ​​4-> 2 before the flow has spent 20)

Therefore, the total flow rate of 20 + 20 + 10 = 50. Output 50.

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answer

• Network flow maximum flow template title

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AC-Code

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 7;
int n, m;

struct Edge {
	int to;
	int next;
	int val;
}edge[maxn << 1]; // 双向边，开 2 倍数组
int head[maxn];
int cnt; // 边的数量，从 0 开始编号
int depth[maxn]; // 分层图标记深度
int cur[maxn]; // 当前弧优化，记录当前点 u 循环到了哪一条边
void add(int u, int v, int w) {
	edge[cnt].to = v;
	edge[cnt].next = head[u];
	edge[cnt].val = w;
	head[u] = cnt++;
}

// bfs分层图
bool bfs(int s, int t) {
	queue<int>q;
	memset(depth, 0, sizeof depth);
	depth[s] = 1; // 源点深度为 1
	cur[s] = head[s];
	q.push(s);
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		for (int i = head[u]; ~i; i = edge[i].next) {
			int v = edge[i].to;
			if (edge[i].val > 0 && depth[v] == 0) { // 如果残量不为0，且 v 还未分配深度
				depth[v] = depth[u] + 1;
				cur[v] = head[v]; //------当前弧优化，注意在bfs里，这样做到了“按需赋值”，因为Dinic本来上界就松得一匹, BFS的过程中不连通的点根本就不用再管了...
				if (v == t) // -----分层图汇点优化：遇到汇点直接返回------
					return true;
				q.push(v);
			}
		}
	}
	return depth[t]; // 汇点深度为 0：不存在分层图，返回false；
					 //           非 0 ：存在增广路，返回true
}

// dfs寻找增广路
int dfs(int u, int flow, int t) {
	if (u == t || flow <= 0) // 到达汇点
		return flow;
	int rest = flow;
	for (int i = cur[u]; ~i; i = edge[i].next) {
		int v = edge[i].to;
		if (depth[v] == depth[u] + 1 && edge[i].val != 0) { // 满足分层图、残量>0 两个条件
			int k = dfs(v, min(rest, edge[i].val), t); // 向下增广
			if (k < 0) // ------无用点优化-----
				depth[v] = 0;
			rest -= k;
			edge[i].val -= k; // 正向边减
			edge[i ^ 1].val += k; // 反向边加
			if (rest <= 0) //------剩余量优化：在进行增广的时候，如果该节点已经没有流量，直接退出------
				break;
		}
	}
	return flow - rest; // flow：推送量，rest：淤积量，flow - rest：接受量/成功传递量
}
int Dinic(int s, int t) {
	int ans = 0;
	while (bfs(s, t)) {
		ans += dfs(s, inf, t);
	}
	return ans;
}

int main() {
	ios;
	int s_, t_;
	while (cin >> n >> m >> s_ >> t_) {
		::cnt = 0;
		memset(head, -1, sizeof head);
		for (int i = 1, a, b, c; i <= m; i++) {
			cin >> a >> b >> c;
			add(a, b, c);
			add(b, a, 0);
		}
		cout << Dinic(s_, t_) << endl;
	}
}