POJ 4786 Fibonacci Tree

Fibonacci Tree

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on  HDU. Original ID: 4786
64-bit integer IO format: %I64d      Java class name: Main

 

　　Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do some research on Spanning Tree. So Coach Pang decides to solve the following problem:
　　Consider a bidirectional graph G with N vertices and M edges. All edges are painted into either white or black. Can we find a Spanning Tree with some positive Fibonacci number of white edges?
(Fibonacci number is defined as 1, 2, 3, 5, 8, ... )

 

Input

　　The first line of the input contains an integer T, the number of test cases.
　　For each test case, the first line contains two integers N(1 <= N <= 105) and M(0 <= M <= 105).
　　Then M lines follow, each contains three integers u, v (1 <= u,v <= N, u<> v) and c (0 <= c <= 1), indicating an edge between u and v with a color c (1 for white and 0 for black).

 

Output

　　For each test case, output a line “Case #x: s”. x is the case number and s is either “Yes” or “No” (without quotes) representing the answer to the problem.

 

Sample Input

2

4 4

1 2 1

2 3 1

3 4 1

1 4 0

5 6

1 2 1

1 3 1

1 4 1

1 5 1

3 5 1

4 2 1

Sample Output

Case #1: Yes

Case #2: No

Source

2013 Asia Chengdu Regional Contest

 

解题：最小生成树Kruskal思想。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 100010;

18 struct arc {

19     int u,v,c;

20 };

21 int uf[maxn],n,m,cnt;

22 int fib[50] = {1,2};

23 arc e[maxn];

24 void init() {

25     for(int i = 2; i < 30; i++)

26         fib[i] = fib[i-1] + fib[i-2];

27 }

28 bool cmp1(const arc &x,const arc &y) {

29     return x.c < y.c;

30 }

31 bool cmp2(const arc &x,const arc &y) {

32     return x.c > y.c;

33 }

34 int Find(int x) {

35     if(x == uf[x]) return uf[x];

36     return uf[x] = Find(uf[x]);

37 }

38 int kruskal(bool flag) {

39     for(int i = 0; i <= n; i++) uf[i] = i;

40     int k = cnt = 0;

41     if(flag) sort(e,e+m,cmp1);

42     else sort(e,e+m,cmp2);

43     for(int i = 0; i < m; i++) {

44         int tx = Find(e[i].u);

45         int ty = Find(e[i].v);

46         if(tx != ty) {

47             uf[tx] = ty;

48             if(e[i].c) k++;

49             cnt++;

50         }

51     }

52     return k;

53 }

54 int main() {

55     int t,x,y,cs = 1;

56     init();

57     scanf("%d",&t);

58     while(t--) {

59         scanf("%d %d",&n,&m);

60         for(int i = 0; i < m; i++)

61             scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].c);

62         x = kruskal(true);

63         if(cnt < n-1) {

64             printf("Case #%d: No\n",cs++);

65             continue;

66         }

67         y = kruskal(false);

68         int *tmp = lower_bound(fib,fib+30,x);

69         if(*tmp >= x && *tmp <= y) 

70             printf("Case #%d: Yes\n",cs++);

71         else printf("Case #%d: No\n",cs++);;

72     }

73     return 0;

74 }

View Code