hdu 4786(生成树)

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4028    Accepted Submission(s): 1252

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, ... )

 

 

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

题意：n个点m条边,其中有一些边是白边，一些边是黑边，问是否存在一棵树上的白边数量是斐波拉契数列里面的某个数.

题解：很巧妙的思想,先按白边排序将白边最多的树选出来，然后黑边排序将白边最少的树选出来。然后如果有斐波拉契数在两棵树的大小中间（因为如果存在的话，是可以通过删边得到的），如果存在，就Ok,还要判一下连通分量。

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
const int N = 100005;
int father[N];
struct Edge{
    int u,v,color;
}edge[N];
int _find(int x){
    if(x!=father[x]) {
        father[x] = _find(father[x]);
    }
    return father[x];
}

int n,m;
int cmp(Edge a,Edge b){
    return a.color>b.color;
}
int cmp1(Edge a,Edge b){
    return a.color<b.color;
}
int kruskal(){
    int cost=0;
    for(int i=0;i<m;i++){
        int x=_find(edge[i].u);
        int y=_find(edge[i].v);
        if(x!=y){
            father[x] = y;
            cost+=edge[i].color;
        }
    }
   return cost;
}
bool vis[N];
void init(){
    memset(vis,false,sizeof(vis));
    int a=1,b=2;
    vis[1]=true,vis[2] =true;
    while(a+b<N){
        vis[a+b]=true;
        swap(a,b);
        b = a+b;
    }
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    int t = 1;
    init();
    while(tcase--){
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].color);
        }
        for(int i=1;i<=n;i++) father[i] = i;
        sort(edge,edge+m,cmp);
        int maxn = kruskal();
        int ans = 0;
        for(int i=1;i<=n;i++){
            if(father[i]==i) ans++;
            if(ans>1) break;
        }
        if(ans>1) {
            printf("Case #%d: No\n",t++);
            continue;
        }
        for(int i=1;i<=n;i++) father[i] = i;
        sort(edge,edge+m,cmp1);
        int minn = kruskal();
        ans = 0;
        for(int i=1;i<=n;i++){
            if(father[i]==i) ans++;
            if(ans>1) break;
        }
        if(ans>1) {
            printf("Case #%d: No\n",t++);
            continue;
        }
        //printf("%d %d\n",minn,maxn);
        bool flag = false;
        for(int i=minn;i<=maxn;i++){
            if(vis[i]){
                flag = true;
                break;
            }
        }
        if(flag) printf("Case #%d: Yes\n",t++);
        else printf("Case #%d: No\n",t++);
    }
    return 0;
}