POJ 1733 Parity Game(并查集+数据的离散化)

Parity game
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9098 Accepted: 3522
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' 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
Source

CEOI 1999

题意：一个长为N的01串，给出M条从A到B位的1的个数的奇偶信息，前x条是没有逻辑矛盾的，求x;

做法，看看数据范围根本开不起来数组了，只能用数据的离散化，
给出的A,B,R其实是说明sumB-sumA的奇偶性，sumA,sumB同奇偶得出结果为偶，不同奇偶得出结果则为奇，用偏移量并查集即可
数据的离散化

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define maxn 10010using namespace std;int n,m;struct Node{    int a,b,re;}node[maxn];int pre[maxn];int off[maxn];int fr(int x){    if(x==pre[x])return x;    int r=fr(pre[x]);    off[x]=(off[x]+off[pre[x]])%2;    return pre[x]=r;}int ap[maxn];int main(){    int a,b;    char str[10];        scanf("%d%d",&n,&m);        for(int i=0;i<=m*2;i++)pre[i]=i;        memset(off,0,sizeof(off));        int ans=0;        int cnt=0;        bool flag=true;        for(int i=0;i<m;i++)        {            scanf("%d%d%s",&a,&b,&str);            b=b+1;            ap[cnt++]=a;            ap[cnt++]=b;            node[i].a=a;            node[i].b=b;            node[i].re=str[0]=='e'? 0:1;        }        sort(ap,ap+2*m);        int siz=unique(ap,ap+cnt)-ap;        for(int i=0;i<m;i++)        {            int st=lower_bound(ap,ap+siz,node[i].a)-ap;            int ed=lower_bound(ap,ap+siz,node[i].b)-ap;            int r1=fr(st);            int r2=fr(ed);            if(r1==r2)            {                if((off[st]-off[ed]+2)%2!=node[i].re){ans=i;flag=false;break;}            }            else            {                pre[r1]=r2;                off[r1]=(-off[st]+node[i].re+off[ed])%2;            }        }        if(flag)printf("%d\n",m);        else printf("%d\n",ans);    return 0;}