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

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
",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
",cs++);
71         else printf("Case #%d: No
",cs++);;
72     }
73     return 0;
74 }

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