H - Parity game

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' orodd' (the answer, i.e. the parity of the number of ones in the chosen subsequence, where even' means an even number of ones andodd' 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

题意:
给定01字串的长度,在给定m个询问区间的奇偶性。
若m句话中第k+1是第一次与前面的话矛盾, 输出k;

题解:

因为题目要求的是奇偶性,可知,一个01数列中从第l个元素到第r个元素之间有奇数个1,如果令sum[]为前缀和,则等价于sum[l-1]^sum[r]=1;
偶数个1的情况就是sum[l-1]^sum[r]=0;当然这个题目中我们不需要求出sum[]具体是多少,只需要知道它们之间的关系就行了。

那么我们就可以设置并查集了,设置一个d[]数组,表示x与p[x]之间的关系(也就是sum[x]^sum[p[x]])。每次询问时,先令x=l-1,y=r,找到p[x]与p[y],再判断一下,如果x,y在同一并查集里,那么就直接看d[x]^d[y]是否等于所询问的关系,如果不满足,直接输出答案然后退出;否则,说明目前还不知道x,y之间的关系,那就将x,y合并到同一个并查集。在操作过程中,维护d[]数组即可。
数据要先离散化

代码:

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 10;
int n, m, cnt = 0;
int ans;
int a[maxn];
int p[maxn];
int d[maxn];
char s[5];
struct node{
  int l, r, v;
}t[maxn];

void init(){
  scanf("%d%d",&n,&m);
  for(int i = 1; i <= m; i ++){
    scanf("%d%d%s",&t[i].l,&t[i].r,s);
    t[i].v = (s[0] == 'e' ? 0 : 1);
    a[++ cnt] = t[i].l - 1;
    a[++ cnt] = t[i].r;
  }
  sort(a + 1, a + cnt + 1);
  n = unique(a + 1, a + cnt + 1) - (a + 1);

  for(int i = 1; i <= n; i ++)  p[i] = i;
}

int find(int x){
  if(x != p[x]){
    int t = find(p[x]);
    d[x] ^= d[p[x]];
    p[x] = t;
  }
  return p[x];
}

int main()
{
  init();

  for(int i = 1; i <= m; i ++){
    int x = lower_bound(a + 1, a + n + 1, t[i].l - 1) - a;
    int y = lower_bound(a + 1, a + n + 1, t[i].r) - a;
    int px = find(x);
    int py = find(y);
    if(px == py){
      if(d[x] ^ d[y] != t[i].v){
        printf("%d\n", i - 1);
        return 0;
      }
    }
    else{
      p[px] = py;
      d[px] = d[y] ^ d[x] ^ t[i].v;
    }
  }
  printf("%d\n", m);
  return 0;
}

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