"ZOJ" 1015 Fishing Net

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1015

Test instructions: Gives an n-point-free graph asking whether it is a string, and the string is defined as having an edge connected to an adjacent point on the ring of an arbitrary length >3 in the graph (n<=1000)

#include <bits/stdc++.h>using namespace Std;const int n=1005;int N, M, Ihead[n], CNT, tag[n], Pos[n];bool vis[n];str UCT E {int next, to;} e[n*n];void Add (int u, int v) {E[++cnt]={ihead[u], v}; Ihead[u]=cnt;e[++cnt]={ihead[v], u}; ihead[v]=cnt;} void Cln () {memset (ihead, 0, sizeof (int) * (n+1)), memset (tag, 0, sizeof (int) * (n+1)), memset (POS, 0, sizeof (int) * (n+1)); cnt=0;} BOOL Check () {int x, Y, MN, mx;for (int now=n; now;--now) {mx=-1; mn=~0u>>1; x=y=0;for (int i=1; i<=n; ++i) if (!po S[i] && tag[i]>mx) mx=tag[i], x=i;pos[x]=now;for (int i=ihead[x]; i; i=e[i].next) if (!pos[e[i].to]) tag[e[i]. to]++;for (int i=ihead[x]; i; i=e[i].next) if (pos[e[i].to]>pos[x] && pos[e[i].to]<mn) mn=pos[e[i].to], y= e[i].to;for (int i=ihead[y]; i; i=e[i].next) vis[e[i].to]=1; vis[y]=1;for (int i=ihead[x]; i; i=e[i].next) if (Pos[e[i].to]>pos[x] &&!vis[e[i].to]) return 0;for (int i= Ihead[y]; I I=e[i].next) vis[e[i].to]=0; vis[y]=0;} return 1;} int main () {while (scanf ("%d%d", &n, &m), n&&m) {int x, y;for (int i=0; i<m; ++i) {scanf ("%d%d", &x, &y); Add (x, y);} Check ()? puts ("Perfect"):p UTS ("imperfect"); Puts (""); Cln ();} return 0;}

　　

String Graph judging naked question = = Detailed view of CDQ's thesis = = "Chord and Interval map"

Chord Graph definition above said = =

Here's a look at the nature:

Regiment: A complete picture, in which any point pair in a regiment is connected by a side.

Simple point: If $x$ and its adjacent points make up a regiment, then $x$ is a simple point.

Perfect elimination sequence: a sequence of points, with each point appearing and once, and satisfying for any $v_i$, $v_i$ is a simple point in the $v_{i+1} \cdots v_{n}$ induced sub-graph.

Theorem 1: A chord graph has at least one perfect elimination sequence. (Proof to see thesis)

Theorem 2: The induced sub-graph of a string graph is also a chord graph

So the algorithm of naked searching for perfect sequence is to try to join the perfect sequence of the current maintenance to see if it is a simple point if it is not in the perfect sequence at each time. Of course this is naked violence = =

So CDQ paper introduces two algorithms = = one is dictionary sequence BFS. One is the most popular algorithm, because I see the online implementation are the most popular algorithm (MCS algorithm) = = So I learned the most trend is OK. Anyway, both algorithm complexity is $o (n+m) $

First, the MCS principle is to find a sequence and then determine whether this is a perfect elimination sequence. So how does the MCS algorithm find a sequence?

Ghosts know! Anyway, looks like this is a kind of greedy QAQCDQ paper also does not explain qaq, each time to the perfect sequence before adding a point, and each time the point is added to the current maintenance sequence of points in the most not in the sequence of points = = what the hell!

So the algorithm is to find the points in the sequence with the most points in the series not the point in the sequence, adding = =

Finally determine whether the sequence is legal:

If we want to judge $v_i$ this point, that is, in $v_{i+1} \cdots v_{n}$ to find the $v_i$ adjacent to the point set $v_{j_1}, v_{j_2}, \cdots v_{j_k}$, and is in the order of the sequence from small to large order. Then we only need to judge $v_{j_1}$ and $v_{j_i}, i>1$ whether there is an edge to connect to, if not, this figure is not a string diagram. Since this is added sequentially to the sequence, and we ask that the point set be a regiment, then obviously we just need to determine whether the first $v_{j_1}$ in the sequence is connected to the other, because $v_{j_i}, i>1$ has already been judged = =

At the point of finding the most connected edge that step, is able to use the list to optimize to $o (1) $, but I am too lazy, direct violence = = Anyway the subject can be too ... Even without $o (1) $, we can also use set = =

"ZOJ" 1015 Fishing Net