很像SCOI的游戏(其实基本就是一样)

首先我们可以推出一个性质,当且仅当某一个连通块中没有环存在输出NIE(无解的意思)

因为有环的话。。。就可以构造一棵基环外向树

所以做法就和SCOI的游戏很像了,并查集维护即可。

 

 1 /**************************************************************
 2     Problem: 1116
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:252 ms
 7     Memory:1292 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11  
12 using namespace std;
13 const int N = 100005;
14 int n, m, fa[N];
15 bool vis[N];
16  
17 inline int read(){
18     int x = 0;
19     char ch = getchar();
20     while (ch < '0' || ch > '9')
21         ch = getchar();
22     while (ch >= '0' && ch <= '9'){
23         x = x * 10 + ch - '0';
24         ch = getchar();
25     }
26     return x;
27 }
28  
29 int find_fa(const int x){
30     return x == fa[x] ? x : fa[x] = find_fa(fa[x]);
31 }
32  
33 int main(){
34     int i, x, y;
35     n = read(), m = read();
36     for (i = 1; i <= n; ++i)
37         fa[i] = i;
38     for (i = 1; i <= m; ++i){
39         x = find_fa(read());
40         y = find_fa(read());
41         if (x != y){
42             fa[x] = y;
43             vis[y] |= vis[x];
44         } else vis[x] = 1;
45     }
46     for (i = 1; i <= n; ++i)
47         if (!vis[find_fa(i)]){
48             puts("NIE");
49             return 0;
50         }
51     puts("TAK");
52     return 0;
53 }
View Code

 

By Xs酱~ 转载请说明 博客地址:http://www.cnblogs.com/rausen
posted on 2014-11-04 19:26  Xs酱~  阅读(264)  评论(0编辑  收藏  举报