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;// & 0x7FFFFFFFconst 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;}