题意:
给出你每对点的链接情况,问你最后构成的是不是一棵树。
解析:
并查集。有以下几点需要判断。
1. 空树是一棵树
2. 自环不算树
3. 森林不算树
4. 构成环路的不算树
AC代码
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 1005;
int pa[N], vis[N];
void init() {
for(int i = 0; i < N; i++) {
pa[i] = i;
vis[i] = false;
}
}
int find(int x) {
if(x == pa[x]) return x;
else return pa[x] = find(pa[x]);
}
void Union(int x, int y) {
int r1 = find(x);
int r2 = find(y);
if(r1 == r2) return ;
vis[x] = vis[y] = true;
if(r1 < r2) pa[r2] = r1;
else pa[r1] = r2;
}
bool judge() {
int x, y;
bool tree = true;
while(scanf("%d%d", &x, &y) != EOF && (x || y)) {
if(find(x) == find(y)) tree = false;
Union(x, y);
}
if(!tree) return false;
int cnt = 0;
for(int i = 0; i < N; i++) {
if(vis[i] && i == pa[i]) cnt++;
}
if(cnt != 1) return false;
return true;
}
int main() {
int cas = 1, x, y;
while(scanf("%d%d", &x, &y) != EOF) {
if(x == -1 && y == -1) break;
init();
if(!x && !y) {
printf("Case %d is a tree.\n", cas++);
continue;
}
Union(x, y);
if(judge()) {
printf("Case %d is a tree.\n", cas++);
}else {
printf("Case %d is not a tree.\n", cas++);
}
}
return 0;
}