POJ 1733 Parity game

Parity game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4587		Accepted: 1799

Description

Now and then you play the following game with your friend. Your friend writes down a sequence consisting of zeroes and ones. You choose a continuous subsequence (for example the subsequence from the third to the fifth digit inclusively) and ask him, whether this subsequence contains even or odd number of ones. Your friend answers your question and you can ask him about another subsequence and so on. Your task is to guess the entire sequence of numbers. 


You suspect some of your friend's answers may not be correct and you
 want to convict him of falsehood. Thus you have decided to write a 
program to help you in this matter. The program will receive a series of
 your questions together with the answers you have received from your 
friend. The aim of this program is to find the first answer which is 
provably wrong, i.e. that there exists a sequence satisfying answers to 
all the previous questions, but no such sequence satisfies this answer.

Input

The first line of input contains one number, which is the length of the sequence of zeroes and ones. This length is less or equal to 1000000000. In the second line, there is one positive integer which is the number of questions asked and answers to them. The number of questions and answers is less or equal to 5000. The remaining lines specify questions and answers. Each line contains one question and the answer to this question: two integers (the position of the first and last digit in the chosen subsequence) and one word which is either `even' or `odd' (the answer, i.e. the parity of the number of ones in the chosen subsequence, where `even' means an even number of ones and `odd' means an odd number).

Output

There is only one line in output containing one integer X. Number X says that there exists a sequence of zeroes and ones satisfying first X parity conditions, but there exists none satisfying X+1 conditions. If there exists a sequence of zeroes and ones satisfying all the given conditions, then number X should be the number of all the questions asked.

Sample Input

10
5
1 2 even
3 4 odd
5 6 even
1 6 even
7 10 odd

Sample Output

3

Source

CEOI 1999

//脑袋不好使呀，经验不足
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <cmath>
#define N 5677
using namespace std;
struct node
{
    int val,next;
}ed[N];
int h[N];
int f[N],s[N];
int Find(int x)
{
    if(x!=f[x])
    {
        int tp=Find(f[x]);
        s[x]^=s[f[x]];//想了一下午才明白为什么这样是对的，因为压缩路径后都直接连到根了，不会再有2次
        return f[x]=tp;
    }
    return x;
}
int cnt;
int hash(int n)
{
   int e,m=n%N;
   for(e=h[m];e!=-1;e=ed[e].next)
   {
       if(ed[e].val==n)
         return e;
      if(ed[e].next==-1)
        break;
   }
    ed[cnt].val=n;
    ed[cnt].next=h[m];
    h[m]=cnt++;
    return cnt-1;
}
int xo;
bool un(int x,int y)
{
    int rx=Find(x);
    int ry=Find(y);
    if(rx!=ry)
    {
       f[ry]=rx;
       s[ry]=s[x]^s[y]^xo;
       return false;
    }
    return true;
}
int main()
{
  //  freopen("in.txt","r",stdin);
    int i,n;
    int len;
    char op[6];
    memset(h,-1,sizeof(h));
    int from,to;
    for(i=0;i<=N;i++)
      f[i]=i;
    scanf("%d",&len);
    scanf("%d",&n);
   for(i=0;i<n;i++)
    {
       scanf("%d %d %s",&from,&to,op);
        xo=0;
       if(op[0]=='o')
        xo=1;
       from=hash(from-1);
       to=hash(to);
       if(un(from,to))
       {
            if((s[from]^s[to])!=xo)
            {
               break;
            }
       }
    }
    printf("%d\n",i);
    return 0;
}
//以下是在 http://acm.timus.ru/problemset.aspx
http://acm.timus.ru/problem.aspx?space=1&num=1003
Ac的代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define N 5377
#define M 10002
struct node
{
   int val;
   int next;
}E[N];
int v[N];
int cnt;
int hsh(int n)//这里不能写hash，会和STL里面hash重到
{
    int k=n%N;
    for(int e=v[k];e!=-1;e=E[e].next)
    {
        if(E[e].val==n)
           return e;
    }
    E[cnt].next=v[k];
    E[cnt].val=n;
    v[k]=cnt++;
    return cnt-1;
}
int F[M],r[M];
int Find(int x)
{
    if(x!=F[x])
    {
        int pf=F[x];
        F[x]=Find(F[x]);
        r[x]=r[x]^r[pf];
        return F[x];
    }
    return x;
}
int main()
{
    int len,q;
    int a,b;
    int i,Xor;
    char op[6];
    while(scanf("%d",&len),len!=-1)
    {
      scanf("%d",&q);
      cnt=0;
      int t=q<<1;
      for(i=0;i<t;i++)
       F[i]=i,r[i]=0;
      memset(v,-1,sizeof(v));
      for(i=0;i<q;i++)
      {
          scanf("%d %d %s",&a,&b,op);
          a=hsh(a-1);
          b=hsh(b);
          Xor=0;
          if(op[0]=='o')
          Xor=1;
          int x=Find(a);
          int y=Find(b);
          if(x==y)
          {
              if((r[a]^r[b])!=Xor)
                break;
          }
          else
          {
              F[y]=x;
              r[y]=r[a]^r[b]^Xor;
          }
      }
      printf("%d\n",i);
      if(i<q) i++;
      for(;i<q;i++)
        scanf("%d %d %s",&a,&b,op);
    }
    return 0;
}