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;
}