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 = 100010;
struct arc {
    int u,v,c;
};
int uf[maxn],n,m,cnt;
int fib[50] = {1,2};
arc e[maxn];
void init() {
    for(int i = 2; i < 30; i++)
        fib[i] = fib[i-1] + fib[i-2];
}
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 = 0; i <= n; i++) uf[i] = i;
    int k = cnt = 0;
    if(flag) sort(e,e+m,cmp1);
    else sort(e,e+m,cmp2);
    for(int i = 0; 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 = 1;
    init();
    scanf("%d",&t);
    while(t--) {
        scanf("%d %d",&n,&m);
        for(int i = 0; i < m; i++)
            scanf("%d %d %d",&e[i].u,&e[i].v,&e[i].c);
        x = kruskal(true);
        if(cnt < n-1) {
            printf("Case #%d: No
",cs++);
            continue;
        }
        y = kruskal(false);
        int *tmp = lower_bound(fib,fib+30,x);
        if(*tmp >= x && *tmp <= y) 
            printf("Case #%d: Yes
",cs++);
        else printf("Case #%d: No
",cs++);;
    }
    return 0;
}