Codeforces Round #396 (Div. 2)D. Mahmoud and a Dictionary（带权并查集）

题目链接：点击打开链接

思路：

带权并查集水题。  带权并查集可以知道在一个集合里的两点间距离。那么这种同义反义关心恰好对应距离的奇偶。

附上一图：

这就是合并的过程。

细节参见代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <ctime>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e-6;
const double PI = acos(-1);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 1e5+10;
int T,n,m,q,p[maxn],dist[maxn];
map<string, int> mp;
char s[33], s1[33];
int _find(int x) {
    if(p[x] == x) return x;
    int oldfa = p[x];
    p[x] = _find(p[x]);
    dist[x] = (dist[x] + dist[oldfa])%2;
    return p[x];
}

void init() {
    for(int i = 1; i <= n; i++) {
        p[i] = i;
        dist[i] = 0;
    }
}

int main() {
    scanf("%d%d%d", &n, &m, &q);
    for(int i = 1; i <= n; i++) {
        scanf("%s", s);
        mp[s] = i;
    }
    init();
    for(int i = 1; i <= m; i++) {
        int id; scanf("%d%s%s", &id, s, s1);
        int id1 = mp[s], id2 = mp[s1];
        int x = _find(id1), y = _find(id2);
        if(x != y) {
            printf("YES\n");
            if(id == 1) p[x] = y, dist[x] = (dist[id2]-dist[id1]+2)%2;
            else p[x] = y, dist[x] = (dist[id2]-dist[id1]+1+2)%2;
        }
        else {
            int cur = (dist[id1]-dist[id2]+2)%2;
            if(id == 1) {
                if(cur & 1) printf("NO\n");
                else printf("YES\n");
            }
            else {
                if(cur & 1) printf("YES\n");
                else printf("NO\n");
            }
        }
    }
    while(q--) {
        scanf("%s%s", s, s1);
        int id1 = mp[s], id2 = mp[s1];
        int x = _find(id1), y = _find(id2);
        if(x != y) printf("3\n");
        else {
            int cur = (dist[id1] - dist[id2] + 2)%2;
            if(cur & 1) printf("2\n");
            else printf("1\n");
        }
    }
    return 0;
}