POJ 1733 Parity Game

Parity game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2426 Accepted: 992

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

1051 2 even3 4 odd5 6 even1 6 even7 10 odd

Sample Output

3

Source

CEOI 1999

 

/*并查集的高级应用对于每一个输入 p1 p2 b,我们可以这样考虑: b = (p2 到某个位置之间的1的个数) - (p1 - 1 到同样某个位置之间的1的个数)的奇偶性。这个位置即为p1 - 1和p2共同的参照位置(并查集的主节点)，因此联想到了并查集，并查集的特点就是处理集合之间的所属关系以及数量关系，具体处理方式是：1)sets[i].sid表示i位置的参照位置，sets[i].dist表示i位置与sets[i].sid位置之间的1的个数的奇偶性2)由于输入num的范围非常大，因此利用Map来做哈希3)奇偶性的处理方式可以用异或来做*/#include <iostream>#include <cstring>#include <map>#define MAX_N 10010using namespace std;struct seg{ //rank是秩，用来加速处理速度 int sid, num, rank; bool dist;}sets[MAX_N + 1];int segNum;char temp[10];map<int, int> hash;void swap(int &v1, int &v2){ int temp = v1; v1 = v2; v2 = temp;}//得到num这个数对应的idint getIndex(int num){ map<int, int>::iterator iter; if((iter = hash.find(num)) != hash.end()) return iter->second; else return -1;}//并查集的find操作int find(int index){ //自身就是当前集合的主节点 if(sets[index].sid == index) return index; else { int pre = sets[index].sid; //更新主节点 sets[index].sid = find(sets[index].sid); //更新相对于主节点的dist sets[index].dist = sets[index].dist ^ sets[pre].dist; return sets[index].sid; }}//处理输入数据num1 num2 distbool joint(int num1, int num2, bool dist){ //得到数据在map中对应的Index int index1 = getIndex(num1), index2 = getIndex(num2); //找到两个元素的主节点id int sid1 = find(index1); int sid2 = find(index2); //相等的话表面num1, num2早已经处理过了，这时需要核对他们之间的距离是不是dist if(sid1 == sid2) return ((sets[index1].dist ^ sets[index2].dist) == dist); //否则对两个集合合并，并更新其中一个集合的主id，这里用到了秩关系来加速处理 if(sets[sid1].rank < sets[sid2].rank) { //更新主节点id sets[sid1].sid = sid2; //更新距离 sets[sid1].dist = sets[index1].dist ^ sets[index2].dist ^ dist; } else { sets[sid2].sid = sid1; sets[sid2].dist = sets[index2].dist ^ sets[index1].dist ^ dist; if(sets[sid1].rank == sets[sid2].rank) sets[sid1].rank++; } return true;}int main(){ int n, sn, i, f, t; bool b; scanf("%d%d", &n, &sn); segNum = 0; int endPos = sn; bool can = true; memset(sets, 0, sizeof(sets)); for(i = 1; i <= sn; i++) { scanf("%d%d%s", &f, &t, temp); if(f > t) swap(f, t); int index1 = getIndex(f - 1); if(index1 == -1) { sets[segNum].num = f - 1; sets[segNum].dist = 0; sets[segNum].sid = segNum; hash[f - 1] = segNum; segNum++; } int index2 = getIndex(t); if(index2 == -1) { sets[segNum].num = t; sets[segNum].dist = 0; sets[segNum].sid = segNum; hash[t] = segNum; segNum++; } if(strcmp(temp, "odd") == 0) b = 1; else b = 0; if(can) { if(!joint(f - 1, t, b)) { endPos = i - 1; can = false; &;nbsp; } } } printf("%d/n", endPos); return 0;}