cf976f Minimal k-covering

枚举 (k),对于每个点 (i) 我们最多删 (deg_i-k) 条边,就源点向第一部、第二部向汇点连边,容量是 (deg_i-k),原边连上,容量是 (1),这样每流过一条原边在网络流图中的边时,就代表这条边可以删掉。也即没有流过的边就是 (k) 时的答案

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
int nu, nv, n, m, ss, tt, hea[4005], deg[4005], cnt, maxFlow, lev[4005], cur[4005];
const int oo=0x3f3f3f3f;
queue<int> d;
vector<int> vec[4005];
struct Edge{
	int too, nxt, val;
}edge[50005], orz[2005];
void add_edge(int fro, int too, int val){
	edge[cnt].nxt = hea[fro];
	edge[cnt].too = too;
	edge[cnt].val = val;
	hea[fro] = cnt++;
}
void addEdge(int fro, int too, int val){
	add_edge(fro, too, val);
	add_edge(too, fro, 0);
}
bool bfs(){
	memset(lev, 0, sizeof(lev));
	lev[ss] = 1;
	d.push(ss);
	while(!d.empty()){
		int x=d.front();
		d.pop();
		for(int i=hea[x]; i!=-1; i=edge[i].nxt){
			int t=edge[i].too;
			if(!lev[t] && edge[i].val>0){
				lev[t] = lev[x] + 1;
				d.push(t);
			}
		}
	}
	return lev[tt]!=0;
}
int dfs(int x, int lim){
	if(x==tt)	return lim;
	int addFlow=0;
	for(int &i=cur[x]; i!=-1; i=edge[i].nxt){
		int t=edge[i].too;
		if(lev[t]==lev[x]+1 && edge[i].val){
			int tmp=dfs(t, min(lim-addFlow, edge[i].val));
			edge[i].val -= tmp;
			edge[i^1].val += tmp;
			addFlow += tmp;
			if(addFlow==lim)	break;
		}
	}
	return addFlow;
}
void dinic(){
	while(bfs()){
		for(int i=ss; i<=tt; i++)	cur[i] = hea[i];
		maxFlow += dfs(ss, oo);
	}
}
int main(){
	memset(hea, -1, sizeof(hea));
	cin>>nu>>nv>>m;
	n = nu + nv;
	ss = 0; tt = n + 1;
	int minDeg=0x3f3f3f3f;
	for(int i=1; i<=m; i++){
		scanf("%d %d", &orz[i].too, &orz[i].nxt);
		orz[i].nxt += nu;
		deg[orz[i].too]++;
		deg[orz[i].nxt]++;
	}
	for(int i=1; i<=n; i++)
		minDeg = min(minDeg, deg[i]);
	for(int i=1; i<=nu; i++)
		addEdge(ss, i, deg[i]-minDeg-1);
	for(int i=1; i<=nv; i++)
		addEdge(nu+i, tt, deg[nu+i]-minDeg-1);
	int tmp=cnt;
	for(int i=1; i<=m; i++)
		addEdge(orz[i].too, orz[i].nxt, 1);
	for(int i=minDeg; i>=0; i--){
		for(int j=0; j<tmp; j+=2)
			edge[j].val++;
		dinic();
		for(int j=tmp; j<cnt; j+=2)
			if(edge[j].val)
				vec[i].push_back((j-tmp)/2+1);
	}
	for(int i=0; i<=minDeg; i++){
		printf("%d ", vec[i].size());
		for(int j=0; j<vec[i].size(); j++)
			printf("%d ", vec[i][j]);
		printf("
");
	}
	return 0;
}
原文地址:https://www.cnblogs.com/poorpool/p/8988818.html