poj 1733 Parity game

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思路: 带权并查集
分析:
1 题目给定n个条件，要我们找到第一个不满足条件的编号
2 每一个条件给的是区间[l,r]的1的奇偶数，很明显[l,r]这个区间的1的个数可以由[0,r]-[0,l-1]得来
3 那么我们利用并查集的思想，rank[x]表示的是x到跟节点这个区间即[x,father[x]]这个区间的1的个数，那么奇偶性可以由1和0来表示
4 题目还有一个难点就是要离散化，一般的离散化的步骤是排序+去重，然后找的时候利用二分即可

代码:

 

#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;

}