HDU-4786-Fibonacci Tree

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HDU-4786-Fibonacci Tree

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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

题目链接：HDU-4786

题目思路：最小生成树，求出1最少的情况a和最多的情况b，在判断在在[a,b]区间是否存在至少一个斐波那契数列的值。

以下是代码：

#include <vector>#include <map>#include <set>#include <algorithm>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>using namespace std;struct node{    int l,r;    int c;  }p[100010];int fa[100010];int n,m;bool cmp1(node a,node b){    if (a.c != b.c)        return a.c < b.c;    return a.l < b.l;}bool cmp2(node a,node b){    if (a.c != b.c)        return a.c > b.c;    return a.l < b.l;}int find(int x)     //并查集{    return fa[x] == x ? x : fa[x] = find(fa[x]);}int solve()    //最小生成树{    for (int i = 0; i <= n; i++)    fa[i] = i;    int cnt = 0;    for (int i = 0; i < m; i++)    {        int u = find(fa[p[i].l]);        int v = find(fa[p[i].r]);        if (p[i].c == 0)        {            if (u != v)            {                fa[v] = u;            }                       }        else        {            if (u != v)            {                fa[v] = u;                cnt++;            }                       }    }    return cnt; }int main(){    int t;    scanf("%d",&t);    long long f[105] = {0};    f[0] = 1;    f[1] = 1;    f[2] = 2;    for (int i = 3; i < 100; i++)    {        f[i] = f[i - 1] + f[i - 2];         }    int ret = 1;    while(t--)    {        scanf("%d%d",&n,&m);        for (int i = 0; i < m; i++)        {            scanf("%d%d%d",&p[i].l,&p[i].r,&p[i].c);        }        sort(p,p + m,cmp1);        int cnt1 = solve();    // 1最少的情况，即先把0全部排完，再接入1        int cou1 = 0;        for (int i = 1;i <= n; i++)  //判断是否连通        {            if (fa[i] == i)                cou1++;        }           sort(p,p + m,cmp2);        int cnt2 = solve();   //1最多的情况，先接1再补充0的情况        int cou2 = 0;        for (int i = 1;i <= n; i++)        {            if (fa[i] == i)                cou2++;        }        if (cou2 != 1 || cou1 != 1)        {            printf("Case #%d: No\n",ret++);        }        else        {            int flag = 0;            for (int i = 0; i < 50; i++)  //判断在[cnt1,cnt2]范围内是否存在至少一个斐波那契的数            {                if (f[i] >= cnt1 && f[i] <= cnt2)                {                    printf("Case #%d: Yes\n",ret++);                    flag = 1;                    break;                }                if (f[i] > cnt2)                    break;            }            if (!flag)                printf("Case #%d: No\n",ret++);        }    }            return 0;}