HDU

题目链接<-点击

　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

没想到这题居然是区域赛的题，应该是一个签到题吧

题目大意，就是有两种颜色的边，黑边和白边，你要把所有的点连接起来，并使得你用的白边树是斐波那契数列中的一个数。

思路：
先打表斐波那契，然后求出来最大生成树（尽可能使用白边）和最小生成树（尽可能使用黑边），若最大值和最小值之间有斐波那契数列中的一个数，就正确，否则就没有。

代码如下：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1000000+100;int vis[maxn],post[maxn];int pre[maxn];int n,m;struct node{    int from,to,val;}road[maxn];int cmp(node aa,node bb){    return aa.val>bb.val;}int cmp1(node aa,node bb){    return aa.val<bb.val;}int findd(int x){    int r=x;    while(r!=pre[r])        r=pre[r];    int i=x,j;    while(i!=r)    {        j=pre[i];        pre[i]=r;        i=j;    }    return r;}int zyz(){    int i;    for(i=0;i<=n;i++)    {        pre[i]=i;    }    sort(road+1,road+1+m,cmp);    int ans=0;    int aaaa=1;    for(i=1;i<=m;i++)    {        int tx=findd(road[i].from);        int ty=findd(road[i].to);        if(tx!=ty)        {            pre[ty]=tx;            ans+=road[i].val;            aaaa++;         }    }    if(aaaa<n)        return -1;    else            return ans;}int zhang(){    int i;    for(i=0;i<=n;i++)    {        pre[i]=i;    }    sort(road+1,road+1+m,cmp1);    int ans=0;    int aaaa=1;    for(i=1;i<=m;i++)    {        int tx=findd(road[i].from);        int ty=findd(road[i].to);        if(tx!=ty)        {            pre[ty]=tx;            ans+=road[i].val;            aaaa++;         }    }    if(aaaa<n)//注意啊！！！！wa了四发，这里以前从来没注意过！！！        return -1;    else        return ans;}int main (){    int t;    memset(post,0,sizeof(post));    memset(vis,0,sizeof(vis));     vis[1]=1;    vis[2]=1;    post[0]=1;    post[1]=2;    int cnt=1;    while(post[cnt]<maxn)    {        post[++cnt]=post[cnt-1]+post[cnt-2];        if(post[cnt]<maxn)        vis[post[cnt]]=1;    }    scanf("%d",&t);    int kk;    for(kk=1;kk<=t;kk++)    {        scanf("%d%d",&n,&m);        int i;        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&road[i].from,&road[i].to,&road[i].val);        }        int ed=zyz();        int st=zhang();        int flag=0;        if(ed==-1||st==-1)        {            printf("Case #%d: No\n",kk);            continue;        }        for(i=st;i<=ed;i++)        {            if(vis[i])            {                flag=1;                break;            }        }        if(flag)            printf("Case #%d: Yes\n",kk);        else            printf("Case #%d: No\n",kk);    }}