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经典的最小路径覆盖问题

可以看出一开始有 1~n条路径(1~n)

每次可以挑选一个点,合并两条路径

但是一个点只能用一次

所以我们把点拆成两个

从源点向每个点的入点连一条边权为1的边

从每个点的出点与汇点连一条边权为1的边

每个点的入点向能到达的点的出点连一条边权为1的边

用 n-最大流 就是答案

至于输出:

突然有点尴尬

随便乱搞吧

用过的边就是合并过的

把这些边合并起来就可以得出路径

(搞个并茶几???)

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;i++)
    #define Fdr(i,x,y) for(int i=x;i>=y;i--)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 355
#define M 6011
int n,head[N],d[N],q[M],cnt=1,st,ed,m;
struct edge{
    int to,nxt,w;
}e[M*2];
bool v[M*2];
void add(int x,int y,int w){
    e[++cnt].to=y;e[cnt].w=w;e[cnt].nxt=head[x];head[x]=cnt;
    e[++cnt].to=x;e[cnt].w=0;e[cnt].nxt=head[y];head[y]=cnt;
}
bool bfs(){
    int h=0,t=1;
    clr(d,0);
    d[q[h]=st]=1;    
    while(h<t){
        int x=q[h++];
        if(x==ed)return 1;
        fl(i,x)if(e[i].w&&!d[to]){
            d[to]=d[x]+1;
            q[t++]=to;
        }
    }
    return 0;
}
int dfs(int x,int mf){
    if(x==ed)return mf;
    int us=0,w;
    fl(i,x)
    if(e[i].w&&d[to]==d[x]+1){
        w=dfs(to,MIN(mf-us,e[i].w));
        e[i].w-=w;e[i^1].w+=w;
        us+=w;
        if(us==mf)return mf;
    }
    if(!us)d[x]=-1;
    return us;
}
int f[N];
int gf(int x){return (f[x]==x)?x:(f[x]=gf(f[x]));}
void op(int x){
    out<<x<<" ";
    fl(i,x)
    if(!e[i].w&&e[i].to>n&&e[i].to<st)op(e[i].to-n);
}
int dinic(){
    int ans=0;
    while(bfs())ans+=dfs(st,inf);
    return ans;
}
int main(){
    in>>n>>m;
    int x,y;
    st=n*2+1,ed=n*2+2;
    Fur(i,1,n)add(st,i,1),add(i+n,ed,1),f[i]=i;
    Fur(i,1,m)
        in>>x>>y,
        add(x,y+n,1);

    int tmp=dinic();
    Fur(i,2,cnt)
        if(e[i^1].to<=n&&e[i].to>n&&e[i].to<st&&!e[i].w)
        f[gf(e[i].to-n)]=gf(e[i^1].to);

    Fur(i,1,n)
    if(gf(i)==i)op(i),out<<"\n";
    out<<n-tmp<<ln;
}
LG 2764 最小路径覆盖问题
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