POJ 1733 Parity game (disjoint-set extension field)

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Meaning of the questions:

Entry $n$ represents a length $n$ the $0,1$ string, $m$ denotes Then there $m$ line input, the next $m$ row input $x, y, even$ represents $x$ to the first $Y$ the middle of characters $1$ number is an even number; $x, y, odd$ represents $x$ to the first $Y$ the middle of characters $1$ is odd number. If the $m$ first sentence $k+1$ is a front for the first time with the words of contradiction, the output $k$。

Analysis of ideas:

$x$ represents $x$ is even in this condition, $x+n$ represents $x$ is an odd condition that

$x, y$ is the same

1. in case $x$ is odd, $Y$ is an odd number
2. in case $x$ is even, $Y$ is an even number

$x, y$ is heterogeneous

1. in case $x$ is even, $Y$ is an odd number
2. in case $x$ is odd, $Y$ is an even number

understanding:

Sideband rights disjoint-set: $x$ and $Y$ are related, they are in the same collection, as to what the relationship is depends on how much weight

Extension field of disjoint-set: the collection as long as a condition is true, then all remaining conditions established inevitable.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>

using namespace std;
const int N = 40010, base = N / 2;

int n, m;
int p[N];
map<int,int> mp;

int get(int x)          // 在线离散化
{
    if(mp.count(x) == 0)    mp[x] = ++n;
    return mp[x];
}

int find(int x)
{
    if(x != p[x])   p[x] = find(p[x]);
    return p[x];
}

int main()
{
    scanf("%d%d", &n, &m);
    
    n = 0;      // 用于离散化
    
    for(int i = 1; i < N; ++i)  p[i] = i;   // 初始化
    
    int res = m;
    for(int i = 1; i <= m; ++i)
    {
        int a, b;
        char op[6];
        scanf("%d%d%s", &a, &b, op);
        
        if(res != m)    continue;
        
        a = get(a - 1), b = get(b);
        
        if(op[0] == 'e')                  // 同类
        {
            if(find(a + base) == find(b))  // 有矛盾
            {
                res = i - 1;
                break;					// 数据没有输入完，可以break
            }
            p[find(a)] = find(b);
            p[find(a + base)] = find(b + base);
        }
        else                            // 异类
        {
            if(find(a) == find(b))      // 有矛盾
            {
                res = i - 1;
                break;
            }
            p[find(a)] = find(b + base);
            p[find(a + base)] = find(b);
        }
    }
    
    printf("%d\n", res);
    
    return 0;
}