Ural1003 Parity 并查集

这类利用并查集判断条件是否矛盾的问题已经不是第一次碰到了。

这次题意是题目给了你一个区间内数字和的奇偶性，让你判断最早第几个条件与前面的条件矛盾，将问题的区间和转化为两个前缀和之差（这类思想经常用到），如果是偶那么两个前缀和奇偶相同，应放在一个集合，如果为奇应该放在不同集合。

每次读入一个条件的时候，判断一下违反规则的点是否在一个集合中即可。

注意都对的时候输出条件数目。

1003. Parity

Time limit: 2.0 second
Memory limit: 64 MB

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

Input contains a series of tests. The first line of each test contains one number, which is the length of the sequence of zeroes and ones. This length is less or equal to 109. In the second line, there is one non-negative integer which is the number of questions asked and answers to them. The number of questions and answers is less or equal to 5 000. 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). The input is ended with a line containing −1.

Output

Each line of 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	output
10 5 1 2 even 3 4 odd 5 6 even 1 6 even 7 10 odd -1	3

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<map>

using namespace std;

#define MAXN 11111

int hash[MAXN];
int set[2*MAXN];
int n,m;

void init()
{
	for(int i=0;i<MAXN*2;i++)
		set[i]=i;
}

int find(int a)
{
	int root=a,temp;
	while(set[root]!=root)
		root=set[root];
	while(set[a]!=root)
	{
		temp=a;
		a=set[a];
		set[temp]=root;
	}
	return root;
}

bool merge(int a,int b)
{
	int x=find(a);
	int y=find(b);
	if(x==y)
		return false;
	set[x]=y;
	return true;
}

int hashit(int x)
{
    int k=x%MAXN;
    while(hash[k]!=-1&&hash[k]!=x)
        k=(k+10)%MAXN;
    if(hash[k]==-1)
        hash[k]=x;
    return k;
}


int main()
{
    while(~scanf("%d",&n) && n!=-1)
    {
        scanf("%d",&m);
        init();
        memset(hash,-1,sizeof(hash));
        int u,v,ans=-1;
        char str[20];
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%s",&u,&v,str);
            if(ans!=-1) continue;
            if(u>v) swap(u,v);
            u=hashit(u-1),v=hashit(v);
            if(str[0]=='e')
            {
                if(find(u)==find(v+MAXN) ||find(u+MAXN)==find(v))
                    ans=i;
                merge(u,v);
                merge(u+MAXN,v+MAXN);
            }
            else
            {
                if(find(u)==find(v)||find(u+MAXN)==find(v+MAXN))
                    ans=i;
                merge(u+MAXN,v);
                merge(v+MAXN,u);
            }
        }
        printf("%d\n",ans!=-1?ans:m);

    }
    return 0;
}