Parity game

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 7567	Accepted: 2951

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

題目：有一個長度 已知的01串，給出多個條件，[l,r]這個區間中1的個數是奇數還是偶數，問前幾個是正確的，沒有矛盾

[l,r]中1個個數可以表示為sum[r]-sum[l-1]，而題目只要求是奇偶，也就確定了sum[r]與sum[l-1]奇偶是否相同，到了這步就簡單了，轉變為經典模型

範圍有點大，按l-1,r離散化一下

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cmath>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <stack>
#include <map>
using namespace std;
#define esp  1e-8
const double PI = acos(-1.0);
const long long inf = 1000000000;
const long long mod = 10000007;
typedef long long LL;
#pragma comment(linker, "/STACK:1024000000,1024000000") 
//freopen("in.txt","r",stdin); //輸入重定向,輸入資料將從in.txt檔案中讀取
//freopen("out.txt","w",stdout); //輸出重定向,輸出資料將儲存在out.txt檔案中
int f[30005];int v[30005];
int a[30005], b[30005];
char s[30005][10];

set<int>st;
set<int>::iterator it;
map<int, int>mp;
int find(int x)
{
	if (x != f[x])
	{
		int u = f[x];
		f[x] = find(f[x]);
		v[x] = (v[u] + v[x]) % 2;
	}
	return f[x];
}

int main()
{
	int n, m, i, j;
	while (~scanf("%d%d", &n, &m))
	{
		int ans = 0;
		int res = 0;
		st.clear();
		mp.clear();
		for (i = 1; i <= m; ++i)
		{
			scanf("%d%d%s", &a[i], &b[i], s[i]);
			st.insert(a[i]);
			st.insert(b[i]);
		}
		int nn = 0;
		for (it = st.begin(); it != st.end(); ++it)
		{
			mp[*it] = ++nn;
		}
		for (i = 0; i <= nn; ++i)
			f[i] = i;
		memset(v, 0, sizeof(v));
		v[0] = 0;
		
		for (i = 1; i <= m; ++i)
		{
			int fa = find(mp[a[i]] - 1);
			int fb = find(mp[b[i]]);
			if (fa == fb)//如果相同，則已經知道了mp[a[i]] - 1和mp[b[i]]的奇偶，所以直接判斷即可
			{
				if (v[mp[a[i]] - 1] == v[mp[b[i]]] && s[i][0] == 'o')
					break;
				if (v[mp[a[i]] - 1] != v[mp[b[i]]] && s[i][0] == 'e')
					break;
				ans++;
			}
			else
			{
				f[fa] = fb;
				if (s[i][0] == 'o') //如果是奇數，需要加上1
				{
					v[fa] = (v[mp[a[i]] - 1] + v[mp[b[i]]] + 1) % 2;
				}
				else
				{
					v[fa] = (v[mp[a[i]] - 1] + v[mp[b[i]]]) % 2;					
				}
				ans++;
			}
		}
		printf("%d\n", ans);	
	}
}