[ZJOI2011]最小割(最小割树模板)

https://www.luogu.com.cn/problem/P3329

最小割树的用处不仅是做这些裸题,了解这个定理,会对一类问题有更深的思考。

最小割树的实现:
每次取两个点u,v,求它们的割,并在最小割树上给它们连边,权值为这个割。
然后按照S能走到的和能走到T的,分成两类点,继续递归建树。

原图中的两个点的最小割,即为树上边权的最小值。

容易证明最小割<=树上边权的最小值,但是要证明恰好是,比较困难,博主暂时不会。

Code:


#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i <  _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("
")
using namespace std;

const int N = 6005;

int n, m;
int x, y, z;

int fi[N], to[N], nt[N], r[N], tot = 1;

void link(int x, int y, int z) {
	nt[++ tot] = fi[x], to[tot] = y, r[tot] = z, fi[x] = tot;
	nt[++ tot] = fi[y], to[tot] = x, r[tot] = z, fi[y] = tot;
}


int dis[N], d[N], d0, S, T;

int bfs() {
	fo(i, 0, n) dis[i] = 1e9;
	dis[S] = 0, d[d0 = 1] = S;
	for(int i = 1; i <= d0; i ++) {
		int x = d[i];
		for(int j = fi[x]; j; j = nt[j]) if(r[j]) {
			int y = to[j];
			if(dis[y] == 1e9) {
				dis[y] = dis[x] + 1;
				d[++ d0] = y;
			}
		}
	}
	return dis[T] != 1e9;
}

int dg(int x, int flow) {
	if(x == T) return flow;
	int use = 0;
	for(int i = fi[x]; i; i = nt[i]) if(r[i] && dis[x] + 1 == dis[to[i]]) {
		int t = dg(to[i], min(flow - use, r[i]));
		use += t, r[i] -= t, r[i ^ 1] += t;
		if(use == flow) return use;
	}
	return use;
}

int mark[N];

void dfs(int x) {
	if(mark[x]) return;
	mark[x] = 1;
	for(int i = fi[x]; i; i = nt[i]) if(r[i])
		dfs(to[i]);
}

#define V vector<int>
#define pb push_back
#define re resize
#define si size()

int ans[N][N];

void cl() {
	for(int i = 2; i <= tot; i += 2) {
		int s = r[i] + r[i ^ 1];
		r[i] = r[i ^ 1] = s / 2;
	}
}

void dg(V a) {
	if(a.si <= 1) return;
	cl();
	S = a[0], T = a[1];
	int sum = 0;
	while(bfs()) sum += dg(S, 1 << 30);
	fo(i, 1, n) mark[i] = 0;
	dfs(S);
	fo(i, 1, n) if(mark[i])
		fo(j, 1, n) if(!mark[j]) {
			if(sum < ans[i][j])
				ans[i][j] = ans[j][i] = sum;
		}
	V b, c; b.clear(); c.clear();
	ff(i, 0, a.si) if(mark[a[i]])
		b.pb(a[i]); else c.pb(a[i]);
	dg(b); dg(c);
}

void cl_e() {
	fo(i, 1, n) fi[i] = 0;
	tot = 1;
}

int main() {
	int T;
	scanf("%d", &T);
	fo(iT, 1, T) {
		cl_e();
		scanf("%d %d", &n, &m);
		fo(i, 1, m) {
			scanf("%d %d %d", &x, &y, &z);
			link(x, y, z);
		}
		fo(i, 1, n) fo(j, 1, n) ans[i][j] = 1e9;
		V a; a.re(n);
		fo(i, 1, n) a[i - 1] = i;
		dg(a);
		int q, x;
		scanf("%d", &q);
		fo(ii, 1, q) {
			scanf("%d", &x);
			int s = 0;
			fo(i, 1, n) fo(j, i + 1, n)
				s += ans[i][j] <= x;
			pp("%d
", s);
		}
		hh;
	}
}
原文地址:https://www.cnblogs.com/coldchair/p/12636739.html