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 }