题解 舞动的夜晚
是个二分图不可行边的模板，可惜我不会
- 二分图必须边判定：边 \((x, y)\) 流量为1并且在残量网络里，x和y属于不同的强连通分量
- 二分图可行边判定：边 \((x, y)\) 流量为1或者在残量网络里，x和y属于同一个强连通分量
于是这题就求出所有可行边，剩下的就是不可行边
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 200010
#define ll long long
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, m, t, lim;
int cnt[N], head[N], size;
struct edge{int from, to, next;}e[N<<1];
inline void add(int s, int t) {e[++size].to=t; e[size].from=s; e[size].next=head[s]; head[s]=size;}
// 这个暴力是假的，暴力应该用最大流跑
namespace force{
int deg[N];
bool del[N], ans2[N], vis[N];
int check(int s) {
// cout<<"check: "<<s<<endl;
for (int i=1; i<=lim; ++i) deg[i]=cnt[i];
memset(vis, 0, sizeof(bool)*(lim+10));
int ans=0;
queue<int> q;
if (~s) {
// cout<<"del: "<<e[s].from<<' '<<e[s].to<<endl;
vis[e[s].from]=vis[e[s].to]=1;
++ans;
q.push(e[s].from); q.push(e[s].to);
}
for (int i=1; i<=lim; ++i) if (deg[i]==1 && !vis[i]) q.push(i);
while (q.size()) {
int u=q.front(); q.pop();
// cout<<"u: "<<u<<endl;
if (vis[u]) {
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (!vis[v] && --deg[v]==1) q.push(v);
}
continue;
}
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
if (vis[v]) continue;
if (vis[u]) {if (--deg[v]==1) q.push(v);}
else {
vis[u]=vis[v]=1; q.push(v);
++ans;
}
}
}
// cout<<"ans: "<<ans<<endl;
// cout<<"vis: "; for (int i=1; i<=lim; ++i) cout<<vis[i]<<' '; cout<<endl;
// cout<<"deg: "; for (int i=1; i<=lim; ++i) cout<<deg[i]<<' '; cout<<endl;
int dlt=0;
for (int i=1; i<=lim; ++i) if (!vis[i] && deg[i]) ++dlt;
return ans+dlt/2;
}
void solve() {
memset(head, -1, sizeof(head));
for (int i=1,u,v; i<=t; ++i) {
u=read(); v=read()+n;
add(u, v); add(v, u);
++cnt[u], ++cnt[v];
}
int ans=0, maxn=check(-1);
// cout<<"maxn: "<<maxn<<endl;
for (int i=1; i<=t; ++i) {
// del[i]=1;
if (check(i*2)<maxn) ++ans, ans2[i]=1;
// del[i]=0;
}
printf("%d\n", ans);
for (int i=1; i<=t; ++i) if (ans2[i]) printf("%d ", i); printf("\n");
}
}
namespace task{
int head[N], size=1, dep[N], cur[N], S, T, scnt, dfn[N], low[N], bel[N], sta[N], top, tot;
bool vis[N<<1], vis2[N];
struct edge{int from, to, next,val;}e[N<<1];
inline void add(int s, int t, int w) {e[++size].to=t; e[size].from=s; e[size].val=w; e[size].next=head[s]; head[s]=size;}
bool bfs(int s, int t) {
memset(dep, 0, sizeof(dep));
queue<int> q;
dep[s]=1;
cur[s]=head[s];
q.push(s);
int u;
while (q.size()) {
u=q.front(); q.pop();
// cout<<"u: "<<u<<endl;
for (int i=head[u],v; ~i; i=e[i].next) {
v = e[i].to;
// cout<<"v: "<<v<<' '<<e[i].val<<' '<<dep[v]<<endl;
if (e[i].val && !dep[v]) {
dep[v]=dep[u]+1;
cur[v]=head[v];
if (v==t) return 1;
q.push(v);
}
}
}
return 0;
}
int dfs(int u, int in) {
if (u==T || !in) return in;
int rest=in;
for (int i=cur[u],v; ~i; cur[u]=i=e[i].next) {
v = e[i].to;
if (e[i].val && dep[v]==dep[u]+1) {
int tem=dfs(v, min(e[i].val, in));
if (!tem) dep[v]=0;
else {
e[i].val-=tem;
e[i^1].val+=tem;
rest-=tem;
}
if (!rest) break;
}
}
return in-rest;
}
void tarjan(int u) {
// cout<<"tarjan: "<<u<<endl;
dfn[u]=low[u]=++tot;
sta[++top]=u; vis2[u]=1;
for (int i=head[u],v; ~i; i=e[i].next) if (e[i].val) {
v = e[i].to;
// cout<<"v: "<<v<<endl;
if (!dfn[v]) {
tarjan(v);
low[u]=min(low[u], low[v]);
}
else if (vis2[v]) low[u]=min(low[u], dfn[v]);
}
if (dfn[u]==low[u]) {
++scnt;
do {
// cout<<"top: "<<sta[top]<<endl;
bel[sta[top]]=scnt; vis2[sta[top--]]=0;
// cout<<"top2: "<<top<<endl;
} while (sta[top+1]!=u);
}
}
void solve() {
memset(head, -1, sizeof(head));
S=lim+1, T=lim+2;
for (int i=1,u,v; i<=t; ++i) {
u=read(); v=read()+n;
add(u, v, 1); add(v, u, 0);
}
for (int i=1; i<=n; ++i) add(S, i, 1), add(i, S, 0);
for (int i=1; i<=m; ++i) add(n+i, T, 1), add(T, n+i, 0);
int tem=0;
while (bfs(S, T)) tem+=dfs(S, INF);
// cout<<"eval: "<<endl; for (int i=2; i<=size; ++i) cout<<"e: "<<e[i].from<<' '<<e[i].to<<' '<<e[i].val<<endl;
// cout<<"tem: "<<tem<<endl;
for (int i=1; i<=lim+2; ++i) if (!dfn[i]) tarjan(i);
// cout<<"scnt: "<<scnt<<endl;
// cout<<"bel: "; for (int i=1; i<=lim; ++i) cout<<bel[i]<<' '; cout<<endl;
int ans=0;
for (int i=2; i<=t*2; i+=2) if (e[i].val && bel[e[i].from]!=bel[e[i].to]) vis[i]=1, ++ans;
printf("%d\n", ans);
for (int i=2; i<=t*2; i+=2) if (vis[i]) printf("%d ", i/2); printf("\n");
}
}
signed main()
{
freopen("night.in", "r", stdin);
freopen("night.out", "w", stdout);
n=read(); m=read(); t=read(); lim=n+m;
// force::solve();
task::solve();
return 0;
}
