Codeforce 22#div2e

Main Topic

Gives N (2<=n<=10^{5) points, starting from each point you are connected to a point V (a total of n edges)}

Now requires adding the fewest edges so the entire graph is a strong connected graph

Approach Analysis

This problem must not generalize: first of all the strong connected components and then the graph into a DAG (we can easily get the number of edges to add, but add which edges will become very troublesome)

Note One detail: The out of each point must be 1

What are the characteristics?

Starting from a point U DFS iterates through all the points that can be traversed, and at the end of the DFS you must get a ring! Also, because the out of each point is 1, all the points that are traversed can only form a ring! And this ring is still at the end of the path, if this is shrunk to a point, then we have to wait for a "inverted long" tree (there are only nodes from the leaf node to the root of the node, the ring is shrunk to the root)

As shown in the following figure:

We take all the points as starting point DFS once again will get a series of this diagram, of course, there is a special case: Ring! To make it easier to tell, we call them "chunking."

Define a starting point and end point for each, and then follow the line below:

When the entire graph has only one ring, it is impossible to add edges to make it a strong connected graph!

Link adjacent two tiles (the end of block A is connected to the beginning of block B)

Create a reverse edge for points that are not the starting point in the block with a 0 entry

Well, after adding the edges, the entire graph becomes a strong connected graph with the smallest number of plus edges.

At first I found a strong connectivity component, and then find out the degree of 0, into the degree of 0, and the degree of access is 0 of the contraction point, but only over 15 groups of data, and then changed to just over 10 sets of data, do not know where the problem, I hope the great God can point out, my error code:

#include <iostream> #include <stdio.h> #include <string> #include <string.h> #include < Vector>const int maxn=200200,maxm=500400;using namespace std;struct edge{int x,y,next;} e[maxm];int num[maxn];///each Connected component contains the number of points int dfn[maxn],low[maxn],v[maxn],s[maxn],b[maxn],h[maxn];int tot=0,cnt=0,times,t;int n,m;void init () {tot=0    ; memset (h,0,sizeof (h));    void Add (int x,int y) {e[++tot].x=x;    E[tot].y=y;    E[TOT].NEXT=H[X]; H[x]=tot;}    int max (int a,int b) {if (a>b) return A; return b;}    int min (int a,int b) {if (a>b) return B; return A;}    void Tarjan (int x) {int y,i;    times++;    t++;    Dfn[x]=low[x]=times;    V[x]=1;    S[t]=x;        for (i=h[x]; i; i=e[i].next) {y=e[i].y;            if (v[y]==0) {Tarjan (y);        Low[x]=min (Low[x],low[y]);    } if (v[y]==1) low[x]=min (Low[x],dfn[y]);        } if (Dfn[x]==low[x]) {cnt++;          do {y=s[t--];  b[y]=cnt;///belongs to which strong connected component, CNT is also the number of strongly connected components 1-cnt v[y]=2;        num[cnt]++;    } while (y!=x);    }}void solve (int n) {times=0;    t=0;    cnt=0;    memset (dfn,0,sizeof (DFN));    memset (num,0,sizeof (num));    memset (v,0,sizeof (v)); for (int i=1; i<=n; i++) if (!dfn[i]) Tarjan (i);}    int In[maxn],out[maxn];int Uin[maxn],uout[maxn],uiu[maxn];vector<int>hd[maxn];int Main () {int t;    scanf ("%d", &t);    Init ();        for (int i=1; i<=t; i++) {int u;        scanf ("%d", &u);    Add (I,u);    } solve (t);        if (cnt==1) {puts ("0");    return 0;    } memset (In,0,sizeof (in));    Memset (out,0,sizeof (out));    memset (uin,0,sizeof (Uin));    memset (uout,0,sizeof (uout));    memset (uiu,0,sizeof (Uiu));            for (int i=1, i<=t; i++) {for (int j=h[i]; j!=0; j=e[j].next) {int y=e[j].y;            Hd[b[i]].push_back (i);            Hd[b[y]].push_back (y);   if (B[i]==b[y]) continue;         out[b[i]]++;        in[b[y]]++;    }} int k=0,k2=0,k1=0;        for (int i=1; i<=cnt; i++) {if (in[i]==0&&out[i]==0) {uiu[k++]=i;            } else {if (in[i]==0) uin[k1++]=i;        if (out[i]==0) uout[k2++]=i;        }} if (K1&GT;K2) {printf ("%d\n", k1+k*2);        for (int i=0; i<k2; i++) printf ("%d%d\n", hd[uout[i]][0],hd[uin[i]][0]);        for (int i=k2; i<k1; i++) printf ("%d%d\n", hd[uout[0]][0],hd[uin[i]][0]);            for (int i=0; i<k; i++) {printf ("%d%d\n", hd[uin[0]][0],hd[uiu[i]][0]);        printf ("%d%d\n", hd[uiu[i]][0],hd[uin[0]][0]);        }} else {printf ("%d\n", k2+k*2);        for (int i=0; i<k1; i++) printf ("%d%d\n", hd[uout[i]][0],hd[uin[i]][0]);        for (int i=k1; i<k2; i++) printf ("%d%d\n", hd[uout[i]][0],hd[uin[0]][0]);     if (k2!=0)   {for (int i=0; i<k; i++) {printf ("%d%d\n", hd[uout[0]][0],hd[uiu[i]][0]);            printf ("%d%d\n", hd[uiu[i]][0],hd[uout[0]][0]); }} else {for (int i=0; i<k-1; i++) {printf ("%d%d\n", Hd[uiu [I]]                [0],hd[uiu[i+1]][0]);            printf ("%d%d\n", hd[uiu[i+1]][0],hd[uiu[i]][0]); }}} return 0;}

AC code, very short, to learn the code of seniors:

#include <iostream> #include <vector> #include <stdio.h> #include <string.h>using namespace std ; const int Maxn=200020;int in[maxn],scl[maxn];vector <int>e[maxn];vector<int>In;vector<int>Out;    int dfs (int u) {scl[u]=1;    int v=e[u][0];    if (!scl[v]) return Scl[u]=dfs (v); else return scl[u]=u;}    int main () {int n,k,t;        while (~SCANF ("%d", &n)) {memset (in,0,sizeof (in));            for (int i=1;i<=n;i++) {int u;            scanf ("%d", &u);            E[i].push_back (U);        in[u]++;        } k=0;        memset (scl,0,sizeof (SCL));                for (int i=1;i<=n;i++) {if (!in[i]) {k++;                In.push_back (i);            Out.push_back (Dfs (i));        }} t=k;               for (int i=1;i<=n;i++) {if (!scl[i]) {k++;               In.push_back (i);            Out.push_back (Dfs (i));}} if (k==1&&t==0) k=0;        printf ("%d\n", K);    for (int i=0;i<k;i++) printf ("%d%d\n", out[i],in[(i+1)%k]);    }//cout << "Hello world!" << Endl; return 0;}

Codeforce 22#div2e