Fibonacci Tree
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 75 Accepted Submission(s): 38
Problem Description
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, ... )
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).
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
只要白边优先和黑边优先两种顺序做两次最小生成树。
得到白边数量的区间，然后枚举斐波那契数列就可以了。
注意如果一开始是非连通的，输出NO
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2013-11-16 14:14:50 4 File Name :E:2013ACM专题强化训练区域赛2013成都1006.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 21 int f[1000010]; 22 23 struct Edge 24 { 25 int u,v,c; 26 }edge[200010]; 27 int F[200010]; 28 int find(int x) 29 { 30 if(F[x] == -1)return x; 31 return F[x] = find(F[x]); 32 } 33 34 bool cmp1(Edge a,Edge b) 35 { 36 return a.c < b.c; 37 } 38 bool cmp2(Edge a,Edge b) 39 { 40 return a.c > b.c; 41 } 42 int main() 43 { 44 //freopen("in.txt","r",stdin); 45 //freopen("out.txt","w",stdout); 46 int tot = 1; 47 f[0] = 1; 48 f[1] = 2; 49 while(f[tot] <= 1000010) 50 { 51 f[tot+1] = f[tot] + f[tot-1]; 52 tot++; 53 } 54 int T; 55 int iCase = 0; 56 int n,m; 57 scanf("%d",&T); 58 while(T--) 59 { 60 iCase++; 61 scanf("%d%d",&n,&m); 62 for(int i = 0;i < m;i++) 63 scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].c); 64 sort(edge,edge+m,cmp1); 65 memset(F,-1,sizeof(F)); 66 int cnt = 0; 67 for(int i = 0;i < m;i++) 68 { 69 int t1 = find(edge[i].u); 70 int t2 = find(edge[i].v); 71 if(t1 != t2) 72 { 73 F[t1] = t2; 74 if(edge[i].c == 1)cnt++; 75 } 76 } 77 int Low = cnt; 78 memset(F,-1,sizeof(F)); 79 sort(edge,edge+m,cmp2); 80 cnt = 0; 81 for(int i = 0;i < m;i++) 82 { 83 int t1 = find(edge[i].u); 84 int t2 = find(edge[i].v); 85 if(t1 != t2) 86 { 87 F[t1] = t2; 88 if(edge[i].c == 1)cnt++; 89 } 90 } 91 int High = cnt; 92 bool ff = true; 93 for(int i = 1;i <= n;i++) 94 if(find(i) != find(1)) 95 { 96 ff = false; 97 break; 98 } 99 if(!ff) 100 { 101 printf("Case #%d: No ",iCase); 102 continue; 103 } 104 bool flag = false; 105 for(int i = 0;i <= tot;i++) 106 if(f[i] >= Low && f[i] <= High) 107 flag = true; 108 if(flag) 109 printf("Case #%d: Yes ",iCase); 110 else printf("Case #%d: No ",iCase); 111 112 } 113 return 0; 114 }