poj 1308 Is It A Tree?
2017-08-13 21:05
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题目:
AC情况:
思路:
判断是不是树,主要两种情况:1.树根不止有一个。2.树根到某个结点的路径不唯一。
主要算法是并查集。
代码:
AC情况:
思路:
判断是不是树,主要两种情况:1.树根不止有一个。2.树根到某个结点的路径不唯一。
主要算法是并查集。
代码:
#include<iostream> using namespace std; int parent[101]; int flag[101]; void init() { for (int x = 0; x < 101; x++) { parent[x] = x; flag[x] = 0; } } int find(int x) { if (parent[x] != x) parent[x] = find(parent[x]); return parent[x]; } int main() { int a, b,c=0,f,i,ctr; while (scanf_s("%d %d",&a,&b)) { if (a == -1 && b == -1) break; if (a == 0 && b == 0) { //空树也算树 printf("Case %d is a tree.\n", ++c); continue; } init(); flag[a] = 1; flag[b] = 1; f = 1; ctr = -1; //ctr 记录树根数量,把0也算作树根ctr++ do { a = find(a); b = find(b); if (a == b)f = 0; else parent[b] = a; scanf_s("%d %d", &a, &b); flag[a] = 1; flag[b] = 1; } while (a != 0 && b != 0); if (f) { for (i = 0; i < 101; i++) if (flag[i] && parent[i] == i)ctr++; if (ctr == 1)printf("Case %d is a tree.\n", ++c); else printf("Case %d is not a tree.\n", ++c); } else printf("Case %d is not a tree.\n", ++c); } return 0; }
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