Zoj 3316 game

The maximum matching of a general image with a flower tree:

After creating the graph, calculate the maximum number of matches.

If there is a Unicom block that is not perfectly matched, the first hand can go to the point that has not been matched. No matter how the latter goes, it will inevitably go to a matched point, the first hand can follow this staggered path. In the end, there must be no path for the later hand, because if there is still a way to go, this staggered path is an augmented path, and there must be a greater match.

Game Time Limit: 1 second memory limit: 32768 KB

Fire and Lam are addicted to the game of go recently. go is one of the oldest board games. it is rich in strategy despite its simple rules. the game is played by two players who alternately place black and white stones on the vacant intersections of a grid of 19*19 lines. once placed on the board, stones cannot be moved elsewhere, unless they are surrounded and captured by the opponent's stones. the object of the game is to control (surround) a larger portion of the board than the opponent.

Fire thinks it is too easy for him to beat Lam. so he thinks out a new game to play on the board. there are some stones on the board, and we don't need to care about the stones 'color in this new game. fire and Lam take turns to remove one of the stones still on the board. but the Manhattan distance between the removed stone and the opponent's last removed stone must not be greaterL. And the one who can't remove any stone loses the game.

The Manhattan distance between (XI, Yi) and (XJ, YJ) is | Xi-XJ | + | Yi-YJ |.

To show the performance of grace, fire lets Lam play first. In the beginning of the game, lam can choose to remove any stone on the board.

Fire and Lam are clever, so they both use the best strategy to play this game. Now, fire wants to know whether he can make sure to win the game.

Input

There are multiple cases (no more than 30 ).

In each case, the first line is a positive integerN(N<= 361) which indicates the number of stones left on the board. Following areNLines, each contains a pair of IntegersXAndY(0 <=X,Y<= 18), which indicate a stone's location. All pairs are distinct. The last line is an integerL(1 <=L<= 36 ).

There is a blank line between cases.

Ouput

If fire can win the game, output "yes"; otherwise, just output "no ".

Sample Input

20 22 0220 22 04

Sample output

NOYES

Author: Lin, Yue
Source: The 10th Zhejiang University Programming Contest
Submit status

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>using namespace std;const int maxn=500;/*******************************************/struct Edge{  int to,next;}edge[maxn*maxn];int Adj[maxn],Size;void init(){  memset(Adj,-1,sizeof(Adj)); Size=0;}void add_edge(int u,int v){  edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;}/*******************************************/int n;int Match[maxn];int Start,Finish,NewBase;int Father[maxn],Base[maxn];bool InQueue[maxn],InPath[maxn],InBlossom[maxn];int Count;queue<int> q;int FindCommonAncestor(int u,int v){  memset(InPath,false,sizeof(InPath));  while(true)    {      u=Base[u];      InPath[u]=true;      if(u==Start) break;      u=Father[Match[u]];    }  while(true)    {      v=Base[v];      if(InPath[v]) break;      v=Father[Match[v]];    }  return v;}void ResetTrace(int u){  int v;  while(Base[u]!=NewBase)    {      v=Match[u];      InBlossom[Base[u]]=InBlossom[Base[v]]=true;      u=Father[v];      if(Base[u]!=NewBase) Father[u]=v;    }}void BlossomContract(int u,int v){  NewBase=FindCommonAncestor(u,v);  memset(InBlossom,false,sizeof(InBlossom));  ResetTrace(u); ResetTrace(v);  if(Base[u]!=NewBase) Father[u]=v;  if(Base[v]!=NewBase) Father[v]=u;  for(int tu=1;tu<=n;tu++)    {      if(InBlossom[Base[tu]])        {          Base[tu]=NewBase;          if(!InQueue[tu])            {              q.push(tu);              InQueue[tu]=true;            }        }    }}void FindAugmentingPath(){  memset(InQueue,false,sizeof(InQueue));  memset(Father,0,sizeof(Father));  for(int i=1;i<=n;i++) Base[i]=i;  while(!q.empty()) q.pop();  q.push(Start); InQueue[Start]=true;  Finish=0;  while(!q.empty())    {      int u=q.front(); InQueue[u]=false;      q.pop();      for(int i=Adj[u];~i;i=edge[i].next)        {          int v=edge[i].to;          if(Base[u]!=Base[v]&&Match[u]!=v)            {              if(v==Start||(Match[v]>0&&Father[Match[v]]>0))                BlossomContract(u,v);              else if(Father[v]==0)                {                  Father[v]=u;                  if(Match[v]>0)                    {                      q.push(Match[v]);                      InQueue[Match[v]]=true;                    }                  else                    {                      Finish=v;                      return ;                    }                }            }        }    }}void AugmentPath(){  int u,v,w;  u=Finish;  while(u>0)    {      v=Father[u];      w=Match[v];      Match[v]=u;      Match[u]=v;      u=w;    }}void Edmonds(){  memset(Match,0,sizeof(Match));  for(int u=1;u<=n;u++)    {      if(Match[u]==0)        {          Start=u;          FindAugmentingPath();          if(Finish>0) AugmentPath();        }    }}struct Point{  int x,y,id;}p[maxn];int L;int MHD(Point a,Point b){  return abs(a.x-b.x)+abs(a.y-b.y);}int main(){  while(scanf("%d",&n)!=EOF)    {      for(int i=1;i<=n;i++)        {          int x,y;          scanf("%d%d",&x,&y);          p[i]=(Point){x,y,i};        }      scanf("%d",&L);      init();      for(int i=1;i<=n;i++)        {          for(int j=1;j<=n;j++)            {              if(i==j) continue;              if(MHD(p[i],p[j])<=L)                {                  add_edge(i,j);                  add_edge(j,i);                }            }        }      Edmonds();      Count=0;      for(int i=1;i<=n;i++)        {          if(Match[i]) Count++;          else break;        }      //cout<<"--> "<<Count<<endl;      if(Count==n) puts("YES");      else puts("NO");    }  return 0;}

Zoj 3316 game