Poj 1733 Parity game

Idea: weighted and query set
Analysis:
1. Given n conditions, we need to find the first number that does not meet the conditions.
2 each condition gives the parity of 1 in the range [l, r]. Obviously, the number of 1 in the range [l, r] can be [0, r]-[0 L-1]
3. Using the concept of querying sets, rank [x] indicates the number of [x, father [x] from x to 1 of the following node, the parity can be expressed by 1 and 0.
4. Another difficulty is discretization. The general discretization step is sorting + deduplication, And Then binary can be used when searching.

Code:

 

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 50010;struct Node{    int x;    int y;    int mark; };Node node[MAXN];int arr[MAXN] , pos;int father[MAXN] , rank[MAXN] , num;void init(){    sort(arr , arr+pos);    num = unique(arr , arr+pos)-arr;    memset(rank , 0 , sizeof(rank));    for(int i = 0 ; i <= num ; i++)        father[i] = i;}int find(int x){    if(father[x] != x){        int fa = father[x];        father[x] = find(father[x]);        rank[x] = (rank[x]+rank[fa])%2;    }    return father[x];}int search(int x){    int left = 0;    int right = num-1;    while(left <= right){        int mid = (left+right)>>1;        if(arr[mid] == x)            return mid;         else if(arr[mid] < x)            left = mid+1;        else             right = mid-1;    }}int solve(int n){    for(int i = 0 ; i < n ; i++){        int x = search(node[i].x)+1;            int y = search(node[i].y)+1;            int fx = find(x);         int fy = find(y);        int val = node[i].mark;        if(fx == fy){            if((rank[y]-rank[x]+2)%2 != val)                 return i;        }        else{            father[fx] = fy;            rank[fx] = (val+rank[y]-rank[x]+2)%2;        }    }    return n;}int main(){    int len , n;    char str[10];    while(scanf("%d%d" , &len , &n) != EOF){        pos = 0;        for(int i = 0 ; i < n ; i++){             scanf("%d %d %s" , &node[i].x , &node[i].y , str);            node[i].x--;            arr[pos++] = node[i].x;            arr[pos++] = node[i].y;            if(str[0] == 'e')                node[i].mark = 0;            else                node[i].mark = 1;        }        init();        printf("%d\n" , solve(n));    }    return 0;}