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【ybtoj】【并查集】【例题1】【模板】并查集
阅读量:328 次
发布时间:2019-03-04

本文共 546 字,大约阅读时间需要 1 分钟。

【例题1】【模板】并查集


Link


解题思路

真真就是个并查集模板


Code

#include 
#include
using namespace std;int n, m, fa[10010], c, x, y, xx, yy;int find(int x) { if(fa[x] != x) { fa[x] = find(fa[x]); return fa[x]; } return x;}int main() { scanf("%d %d", &n, &m); for (int i = 1; i <= n; i++) fa[i] = i; for (int i = 1; i <= m; i++) { scanf("%d %d %d", &c, &x, &y); if(c == 1) { xx = find(x), yy = find(y); if(xx != yy) fa[xx] = yy; } else { xx = find(x), yy = find(y); if(xx != yy) printf("N\n"); else printf("Y\n"); } }}

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