[2016-03-18] [POJ] [1733] [Parity Game]

• Time: 2016-03-18-09:55:52
• Title Number: [2016-03-18][poj][1733][parity game]
• Question: Given a number of intervals within the range of the number of the and, ask from which sentence start is wrong
• Analysis:
  • Take the right and check the set
  • Interval lengths up to 1000000000 it is not possible to create an array directly, but only 5,000 times is found, that is, at most 5000*2 points, discretization can solve the problem.
    • Relation[i] I is the left end of the interval, fa[i] is the right endpoint of the interval, relation[i] maintenance (i,fa[i]) The parity state of the interval, 0 for even, 1 for odd
    • F (A, B) f (b,c) indicates the state of the interval, then f (a,c) = (f (A, B) + f (b,c))%2;

  
 

   
  

  
#include <map>
   
  

  
#include <cstdio>
   
  

  
using namespace std;
   
  

  
typedef long long LL;
   
  

  
 #define   for   ( x   y   z  )    for   ( int     ( x  ) = ( y  );( x   z   x    
   
  

  
const int maxn = 5000*2 + 100;
   
  

  
int fa[maxn],relation[maxn];
   
  

  
map<LL ,int > m;
   
  

  
void ini(){
   
  

  
 m.clear();
   
  

  
 FOR(i,0,maxn) fa[i] = i;
   
  

  
}
   
  

  
int fnd(int x){
   
  

  
 if(x == fa[x]) return x;
   
  

  
 int tmp = fa[x];
   
  

  
 FA  [      =   FND   ( fa  [ x    
   
  

  
 relation[x] = (relation[x] + relation[tmp])&1;
   
  

  
 return fa[x];
   
  

  
}
   
  

  
int uni(int x,int y,int type){
   
  

  
 int fax = fnd(x),fay = fnd(y);
   
  

  
 if(fax == fay) return 0; 
   
  

  
 fa[fax] = fay;
   
  

  
 relation[fax] = (relation[x] + relation[y] + type)&1;
   
  

  
 return 1;
   
  

  
}
   
  

  

   
  

  
int main(){
   
  

  
 //freopen("in.txt","r",stdin);
   
  

  
 //freopen("out.txt","w",stdout);
   
  

  
 int len,n,cur = 0,ans = 0,flg = 1;
   
  

  
 LL u,v;
   
  

  
 char str[10];
   
  

  
 scanf("%d%d",&len,&n);
   
  

  
 ini(); 
   
  

  
 FOR(i,0,n){
   
  

  
 if(flg){
   
  

  
 scanf("%I64d%I64d%s",&u,&v,str);
   
  

  
 --u;
   
  

  
 if(!m.count(u)) u = m[u] = cur++;
   
  

  
 else u = m[u];
   
  

  
 if(!m.count(v)) v = m[v] = cur++;
   
  

  
 else v = m[v];
   
  

  
 int type = (str[0] == ‘o‘?1:0);
   
  

  
 if(!uni(u,v,type)){
   
  

  
 int t = (relation[ u ] + relation[ v ])&1;
   
  

  
 if(t != type){
   
  

  
 flg = 0;
   
  

  
 continue;
   
  

  
 }
   
  

  
 }
   
  

  
 ++ans;
   
  

  
 }else scanf("%*I64d%*I64d%*s");
   
  

  
 }
   
  

  
 printf("%d\n",ans);
   
  

  

   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-03-18] [POJ] [1733] [Parity Game]