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poj1611 简单题
代码中id记录父节点,sz记录子树规模。一个集合为一棵树。
#include <iostream> #include <cstdio> using namespace std; int id[300005]; int sz[300005]; void add(int a, int b) { int i, j; for (i = a; i != id[i]; i = id[i]); for (j = b; j != id[b]; j = id[j]); if (i == j) return; if (sz[i] < sz[j]) { id[i] = j; sz[j] += sz[i]; } else { id[j] = i; sz[i] += sz[j]; } } int main() { int n, m, cnt, root, temp; while (scanf("%d%d", &n, &m) != EOF) { if (!n) break; for (int i = 0; i < n; ++i) { id[i] = i; sz[i] = 1; } for (int i = 0; i < m; ++i) { scanf("%d", &cnt); scanf("%d", &root); for (int j = 1; j < cnt; ++j) { scanf("%d", &temp); add(root, temp); } } int ans = 0; int rt; for (rt = 0; rt != id[rt]; rt = id[rt]); for (int i = 0; i < n; ++i) { int j; for (j = i; j != id[j]; j = id[j]); if (j == rt) ++ans; } printf("%d\n", ans); } return 0; }
poj 2492
题目也是醉了,看半天没看懂= =#
输入每对a b表示a和b是夫妻,问有没有同性恋= =
把每一次a b放入同一集合,并用rel记录每个节点和它父节点的相对关系。这样同一集合的任意两点间关系就确定了
/********************************************** Memory: 8500 KB Time: 125 MS Language: G++ Result: Accepted ***********************************************/ #include <iostream> #include <cstdio> #include <cstring> using namespace std; int id[2010], sz[2010]; int a[1000005], b[1000005]; int rel[2010]; //和父节点的性别是否一致,一致为0,否则为1 int Scan() { //输入外挂 int res = 0, flag = 0; char ch; if((ch = getchar()) == ‘-‘) flag = 1; else if(ch >= ‘0‘ && ch <= ‘9‘) res = ch - ‘0‘; while((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) res = res * 10 + (ch - ‘0‘); return flag ? -res : res; } void add(int a, int b) { int i, j; int rel_a = 0, rel_b = 0; for (i = a; i != id[i]; i = id[i]) rel_a = (rel_a + rel[i]) % 2; //a和根节点的关系 for (j = b; j != id[j]; j = id[j]) rel_b = (rel_b + rel[j]) % 2; //b和根节点的关系 if (i == j) return ; if (sz[i] <= sz[j]) { //i->j sz[j] += sz[i]; id[i] = j; rel[i] = (rel_a == rel_b) ? 1 : 0; } else { //j->i sz[i] += sz[j]; id[j] = i; rel[j] = (rel_a == rel_b) ? 1 : 0; } } int is_gay(int a, int b) { int i, j; int rel_a = 0, rel_b = 0; for (i = a; i != id[i]; i = id[i]) rel_a = (rel_a + rel[i]) % 2; for (j = b; j != id[j]; j = id[j]) rel_b = (rel_b + rel[j]) % 2; if (i == j && rel_a == rel_b) return 1; return 0; } int main() { int t, m, n; t = Scan(); for (int k = 1; k <= t; ++k) { n = Scan(); m = Scan(); for (int i = 1; i <= n; ++i) { id[i] = i; sz[i] = 1; rel[i] = 0; } for (int i = 0; i < m; ++i) { a[i] = Scan(); b[i] = Scan(); add(a[i], b[i]); } int i; for (i = 0; i < m; ++i) { if (is_gay(a[i], b[i])) break; } printf("Scenario #%d:\n%s\n\n", k, i == m ? "No suspicious bugs found!" : "Suspicious bugs found!"); } return 0; }
总结经验教训:以后多敲两行也不能复制粘贴= =太坑。。。
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原文地址:http://www.cnblogs.com/wenruo/p/4658874.html