【poj1733】Parity game--边带权并查集

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15776		Accepted: 5964

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

翻译：

　　小A和小B玩游戏，小A写了一个长度为N的01序列，小B向小A提出许多问题，每个问题，小B指定两个数L和R，小A回答区间中有奇数个1还是偶数个1，小B发现小A可能在撒谎，向你求助，

求出一个最小的K，满足存在01序列满足1~k组询问，不满足第K加一组询问，M<=10000.

思路：

　　假如我们用sum数组表示1的个数的前缀和，每次回答，

　　1.l~r有偶数个1，等价于sum[l-1]与sum[r]的奇偶性相同。

　　2.l~r有奇数个1, 等价于sum[l-1]与sum[r]的 奇偶性不同。

　　由于序列长度很大，我们可以先离散化一下

scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%s",&e[i].l, &e[i].r, s+1);
		e[i].opt=(s[1]=='o'? 1 : 0);
		b[++cnt]=e[i].l-1;
		b[++cnt]=e[i].r;
	}
	sort(b+1,b+cnt+1);
	int len=unique(b+1 ,b +cnt+1)-b-1;
	for(int i=1;i<=len;i++)fa[i]=i;
 
	int x=lower_bound(b+1,b+len+1,e[i].l-1)-b;
	int y=lower_bound(b+1,b+len+1,e[i].r)-b;

　　之后：

　　我们可以用边带权并查集，边权d[x]为零，表明x到fa[x]的奇偶性相同，为1则不同。在路径压缩时，对x到根的路径上的边权异或运算，即可得出x与树根的奇偶性关系。

　　合并时x与y的奇偶性关系ans=d[x] xor d[y] xor d[p].

　　所以 d[p] = d[x] xor d[y] xor ans

　　

int find(int x)
{
	if(x==fa[x])return fa[x];
	int f=find(fa[x]);
	d[x]^=d[fa[x]];
	return fa[x]=f;
}

　　代码：

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define N 10000000
using namespace std;
int d[N],b[N],n,m,cnt,tot,fa[N];
struct node
{
	int l,r,opt;
}e[N<<1];
char s[10];
int find(int x)
{
	if(x==fa[x])return fa[x];
	int f=find(fa[x]);
	d[x]^=d[fa[x]];
	return fa[x]=f;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%s",&e[i].l, &e[i].r, s+1);
		e[i].opt=(s[1]=='o'? 1 : 0);
		b[++cnt]=e[i].l-1;
		b[++cnt]=e[i].r;
	}
	sort(b+1,b+cnt+1);
	int len=unique(b+1 ,b +cnt+1)-b-1;
	for(int i=1;i<=len;i++)fa[i]=i;
	for(int i=1;i<=m;i++)
	{
		int x=lower_bound(b+1,b+len+1,e[i].l-1)-b;
		int y=lower_bound(b+1,b+len+1,e[i].r)-b;
		int fx=find(x),fy=find(y);
		if(fx==fy)
		{
			if((d[x] ^ d[y])!=e[i].opt)
			{
				printf("%d\n",i-1);
				return 0;
			}
		}
		else
		{
			fa[fx]=fy;
			d[fx]=d[x]^d[y]^e[i].opt;
		}
	}
	printf("%d\n",m);
	return 0;
}