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。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 struct arc {
     int u,v,c;
 };
 int uf[maxn],n,m,cnt;
 int fib[] = {,};
 arc e[maxn];
 void init() {
     for(int i = ; i < ; i++)
         fib[i] = fib[i-] + fib[i-];
 }
 bool cmp1(const arc &x,const arc &y) {
     return x.c < y.c;
 }
 bool cmp2(const arc &x,const arc &y) {
     return x.c > y.c;
 }
 int Find(int x) {
     if(x == uf[x]) return uf[x];
     return uf[x] = Find(uf[x]);
 }
 int kruskal(bool flag) {
     for(int i = ; i <= n; i++) uf[i] = i;
     int k = cnt = ;
     if(flag) sort(e,e+m,cmp1);
     else sort(e,e+m,cmp2);
     for(int i = ; i < m; i++) {
         int tx = Find(e[i].u);
         int ty = Find(e[i].v);
         if(tx != ty) {
             uf[tx] = ty;
             if(e[i].c) k++;
             cnt++;
         }
     }
     return k;
 }
 int main() {
     int t,x,y,cs = ;
     init();
     scanf("%d",&t);
     while(t--) {
         scanf("%d %d",&n,&m);
         for(int i = ; i < m; i++)
             scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].c);
         x = kruskal(true);
         if(cnt < n-) {
             printf("Case #%d: No\n",cs++);
             continue;
         }
         y = kruskal(false);
         int *tmp = lower_bound(fib,fib+,x);
         if(*tmp >= x && *tmp <= y)
             printf("Case #%d: Yes\n",cs++);
         else printf("Case #%d: No\n",cs++);;
     }
     return ;
 }