• 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
    ", ans);
    	print(1);
    } 
    

      

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