poj 1733 Parity game

点击打开poj 1733

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;
}