思路:并查集维护,多一个d数组来表示与当前根节点的关系;
/***************************************** Author :Crazy_AC(JamesQi) Time :2015 File Name : *****************************************/ // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <iomanip> #include <sstream> #include <string> #include <stack> #include <queue> #include <deque> #include <vector> #include <map> #include <math.h> #include <set> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> using namespace std; #define MEM(a,b) memset(a,b,sizeof a) typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> ii; const int inf = 1 << 30; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; inline int Readint(){ char c = getchar(); while(!isdigit(c)) c = getchar(); int x = 0; while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); } return x; } int n; const int maxn = 2e4 + 10; int d[maxn],f[maxn]; int find(int x){ if (x == f[x]) return x; int t = find(f[x]); d[x] += d[f[x]]; f[x] = t; return t; } int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int t; scanf("%d",&t); while(t--){ scanf("%d",&n); for (int i = 0;i <= n;i++) f[i] = i,d[i] = 0; char op[10]; int a,b; while(~scanf("%s",op)){ if (op[0] == 'O') break; if (op[0] == 'E'){ scanf("%d",&a); find(a); printf("%d\n",d[a]); } if (op[0] == 'I'){ scanf("%d%d",&a,&b); int t1 = find(a); int t2 = find(b); f[a] = b; d[a] = abs(a - b) % 1000; } } } return 0; }