Codeforces 343E 最小割树

题意及思路：https://www.cnblogs.com/Yuzao/p/8494024.html

最小割树的实现参考了这篇博客：https://www.cnblogs.com/coder-Uranus/p/9771919.html

代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 210;
const int maxm = 2010;
int head[maxn], Next[maxm], ver[maxm], edge[maxm], d[maxn];
int n, m, s, t, tot, maxflow;
int head1[maxn], Next1[maxm], ver1[maxm], edge1[maxm], tot1;
bool v1[maxm];
int ans = 0;
int mi, mi_id;
int a[maxn], b[maxn];
queue<int> q;
struct Edge {
	int u, v, w;
};
Edge e[maxm];
void add(int x, int y, int z) {
	ver[++tot] = y, edge[tot] = z, Next[tot] = head[x], head[x] = tot;
	ver[++tot] = x, edge[tot] = z, Next[tot] = head[y], head[y] = tot;
}
void add1(int x, int y, int z) {
	ver1[++tot1] = y, edge1[tot1] = z, Next1[tot1] = head1[x], head1[x] = tot1;
	ver1[++tot1] = x, edge1[tot1] = z, Next1[tot1] = head1[y], head1[y] = tot1;
}
void init() {
	memset(head, 0, sizeof(head));
	tot = 1;
	for (int i = 1; i <= m; i++)
		add(e[i].u, e[i].v, e[i].w);
}
bool bfs() {
	memset(d, 0, sizeof(d));
	while(q.size()) q.pop();
	q.push(s); d[s] = 1;
	while(q.size()) {
		int x = q.front(); q.pop();
		for (int i = head[x]; i; i = Next[i]) {
			if(edge[i] && !d[ver[i]]) {
				q.push(ver[i]);
				d[ver[i]] = d[x] + 1;
				if(ver[i] == t) return 1;
			}
		}
	}
	return 0;
}
int dfs(int x, int flow) {
	if(x == t) return flow;
	int rest = flow, k;
	for (int i = head[x]; i && rest; i = Next[i]) {
		if(edge[i] && d[ver[i]] == d[x] + 1) {
			k = dfs(ver[i], min(rest, edge[i]));
			if(!k) d[ver[i]] = 0;
			edge[i] -= k;
			edge[i ^ 1] += k;
			rest -= k;
		}
	}
	return flow - rest;
}
void dinic(int x, int y) {
	s = x, t = y;
	init();
	int flow = 0;
	maxflow = 0;
	while(bfs())
		while(flow = dfs(s, INF)) maxflow += flow;
}
void build(int l, int r) {
	if(l == r) return;
	int x = a[l], y = a[l + 1];
	dinic(x, y);
	int L = l - 1, R = r + 1;
	for (int i = l; i <= r; i++) {
		if(d[a[i]] == 0) b[--R] = a[i];
		else b[++L] = a[i];
	}
	for (int i = l; i <= r; i++)
		a[i] = b[i];
	add1(x, y, maxflow);
	//cout << maxflow << endl;
	ans += maxflow;
	build(l, L);
	build(R, r);
}
void dfs1(int x, int fa) {
	for (int i = head1[x]; i; i = Next1[i]) {
		int y = ver1[i], z = edge1[i];
		if(v1[i]) continue;
		if(y == fa) continue;
		if(z < mi) {
			mi = z;
			mi_id = i;
		}
		dfs1(y, x);
	}
}
void print(int x) {
	mi = INF, mi_id = -1;
	dfs1(x, -1);
	int Mi_id = mi_id;
	if(Mi_id == -1) {
		printf("%d ", x);
		return;
	}
	v1[Mi_id] = v1[Mi_id ^ 1] = 1;
	print(ver1[Mi_id]);
	print(ver1[Mi_id ^ 1]);
}
int main() {
	tot1 = 1;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++) {
		scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
	}
	for (int i = 1; i <= n; i++) a[i] = i;
	build(1, n);
	printf("%d\n", ans);
	print(1);
}

原文地址：https://www.cnblogs.com/pkgunboat/p/10954711.html