AtCoder AGC038F Two Permutations (network flow, minimum cut)

Topic Link

https://atcoder.jp/contests/agc038/tasks/agc038_f

answer

Good question.
First we observed a nature, for alignment \ (P \) , each formed in its rotation point \ (A_i \) is selected from the group \ (I \) or choose \ (P_i \) status must be the same. \ (Q_i \) empathy.
Then converted to minimize \ (A_i = B_i \) position \ (I \) number.
Consider \ (A_i = B_i \) conditions:

(. 1) \ (P_i Q_i = I = \) , then this position useless, \ (A_i B_i = \) must satisfy.

(2) \ (P_i = I, Q_i \ Ne I \) , Provisions \ (A_i = B_i \) , etc.价于\ (B_i = I \) .

(3) \ (P_i \ Ne I, Q_i = I \) , Provisions \ (A_i = B_i \) , etc.价于\ (A_i = I \) .

(. 4) \ (P_i Q_i = \ I NE \) , then \ (A_i = B_i \) is equivalent to \ (A_i \) and \ (B_i \) selected status (that is, \ (I \) or \ ( P_i \) or \ (Q_i \) ) different.

(5) \ (P_i \ Ne Q_i \ Ne I \) , Provisions \ (A_i = B_i \) , etc.价于\ (A_i = i \)且\ (B_i = I \) .

Then we can observe from a set partitioning model: all the points are divided into \ (S, T \) two sets, set \ (A_i = i \) represents \ (A_i \) in \ (S \) set, \ (A_i = P_i \) represents \ (A_i \) in \ (T \) set, \ (B_i = I \) represents \ (I \) in \ (T \) set, \ (B_i = Q_i \) represents \ (I \) in \ (S \) set. Then the conditions described above can be converted to:
for each location \ (I \) :

(1) \ (P_i Q_i = \) , this position is useless, be sure to spend \ (1 \) price.

(2) \ (P_i = I, Q_i \ NE I \) , if the \ (B_i \) in \ (T \) Set takes \ (1 \) price.

(. 3) \ (P_i \ NE I, I Q_i = \) , if the \ (A_i \) in \ (S \) Set takes \ (1 \) cost.

(. 4) \ (P_i Q_i = \ I NE \) , if the \ (A_i \) and \ (B_i \) at different set takes \ (1 \) price.

(. 5) \ (P_i \ NE Q_i \ NE I \) , then if \ (A_i \) in \ (S \) Set and \ (B_i \) in \ (T \) Set takes \ (1 \) of cost.

Thus, the establishment of a point of rotation for each, and then for each \ (I \) discussion classification, it corresponds to the rotation side can be connected.
Since the graph is built out of a bipartite graph, and the edge weights are \ (1 \) , so the time complexity \ (O (n-\ n-sqrt) \) .

After I finished up TLE, and finally found that actually I wrote two years to optimize the current arc has been a fake ...... 38 hanging hair really am speechless ......

Code

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
using namespace std;

const int INF = 1e7;

namespace NetFlow
{
    const int N = 2e5+2;
    const int M = 4e5;
    struct Edge
    {
        int v,w,nxt,rev;
    } e[(M<<1)+3];
    int fe[N+3];
    int te[N+3];
    int dep[N+3];
    int que[N+3];
    int n,en,s,t;
    void addedge(int u,int v,int w)
    {
//      printf("addedge %d %d %d\n",u,v,w);
        en++; e[en].v = v; e[en].w = w;
        e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;
        en++; e[en].v = u; e[en].w = 0;
        e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;
    }
    bool bfs()
    {
        for(int i=1; i<=n; i++) dep[i] = 0;
        int head = 1,tail = 1; que[1] = s; dep[s] = 1;
        while(head<=tail)
        {
            int u = que[head]; head++;
            for(int i=fe[u]; i; i=e[i].nxt)
            {
                int v = e[i].v;
                if(e[i].w>0 && dep[v]==0)
                {
                    dep[v] = dep[u]+1;
                    if(v==t)return true;
                    tail++; que[tail] = v;
                }
            }
        }
        return false;
    }
    int dfs(int u,int cur)
    {
        if(u==t||cur==0) {return cur;}
        int rst = cur;
        for(int &i=te[u]; i; i=e[i].nxt)
        {
            int v = e[i].v;
            if(e[i].w>0 && rst>0 && dep[v]==dep[u]+1)
            {
                int flow = dfs(v,min(rst,e[i].w));
                if(flow>0)
                {
                    e[i].w -= flow; 
                    rst -= flow;
                    e[e[i].rev].w += flow;
                    if(rst==0) {return cur;}
                }
            }
        }
        if(rst==cur) {dep[u] = -2;}
        return cur-rst;
    }
    int dinic(int _n,int _s,int _t)
    {
        n = _n,s = _s,t = _t;
        int ret = 0;
        while(bfs())
        {
            for(int i=1; i<=n; i++) te[i] = fe[i];
            memcpy(te,fe,sizeof(int)*(n+1));
            ret += dfs(s,INF);
        }
        return ret;
    }
}
using NetFlow::addedge;
using NetFlow::dinic;

const int N = 1e5;
int a[N+3],b[N+3];
int ca[N+3],cb[N+3];
int n,cnta,cntb;

int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++) scanf("%d",&a[i]),a[i]++;
    for(int i=1; i<=n; i++) scanf("%d",&b[i]),b[i]++;
    for(int i=1; i<=n; i++)
    {
        if(!ca[i])
        {
            cnta++;
            int x = i;
            do
            {
                ca[x] = cnta;
                x = a[x];
            } while(x!=i);
        }
    }
    for(int i=1; i<=n; i++)
    {
        if(!cb[i])
        {
            cntb++;
            int x = i;
            do
            {
                cb[x] = cntb;
                x = b[x];
            } while(x!=i);
        }
    }
//  printf("ca: "); for(int i=1; i<=n; i++) printf("%d ",ca[i]); puts("");
//  printf("cb: "); for(int i=1; i<=n; i++) printf("%d ",cb[i]); puts("");
    int ans = n;
    for(int i=1; i<=n; i++)
    {
        if(a[i]==i && b[i]==i) {ans--;}
        else if(a[i]==i) {addedge(1,cb[i]+cnta+2,1);}
        else if(b[i]==i) {addedge(ca[i]+2,2,1);}
        else
        {
            addedge(ca[i]+2,cb[i]+cnta+2,1);
            if(a[i]==b[i]) {addedge(cb[i]+cnta+2,ca[i]+2,1);}
        }
    }
    ans -= dinic(2+cnta+cntb,1,2);
    printf("%d\n",ans);
    return 0;
}