Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5934 Accepted Submission(s):
1845
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, ... )
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).
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
Recommend
和昨天ysy讲的那道题差不多
而且这道题在题目中直接给提示了——》黑边为0，白边为1
这样的话我们做一个最小生成树和一个最大生成树
如果在这两个值的范围内有斐波那契数，就说明满足条件
简单证明：
对于最小生成树来说，任意删除一条边，并加入一条没有出现过的边，这样的话权值至多加1，边界为最大生成树
对于最小生成树来说，任意删除一条边，并加入一条没有出现过的边，这样的话权值至多加1，边界为最大生成树
#include<cstdio> #include<algorithm> using namespace std; const int MAXN=1e6+10,INF=1e9+10; inline char nc() { static char buf[MAXN],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++; } inline int read() { char c=nc();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();} return x*f; } struct node { int u,v,w; }edge[MAXN]; int num=1; inline void AddEdge(int x,int y,int z) { edge[num].u=x; edge[num].v=y; edge[num].w=z;num++; } int N,M; int fib[MAXN]; int fa[MAXN]; int comp1(const node &a,const node &b){return a.w<b.w;} int comp2(const node &a,const node &b){return a.w>b.w;} int find(int x) { if(fa[x]==x) return fa[x]; else return fa[x]=find(fa[x]); } void unionn(int x,int y) { int fx=find(x); int fy=find(y); fa[fx]=fy; } int Kruskal(int opt) { if(opt==1) sort(edge+1,edge+num,comp1); else sort(edge+1,edge+num,comp2); int ans=0,tot=0; for(int i=1;i<=num-1;i++) { int x=edge[i].u,y=edge[i].v,z=edge[i].w; if(find(x) == find(y)) continue; unionn(x,y); tot++; ans=ans+z; if(tot==N-1) return ans; } } int main() { #ifdef WIN32 freopen("a.in","r",stdin); #else #endif int Test=read(); fib[1]=1;fib[2]=2; for(int i=3;i<=66;i++) fib[i]=fib[i-1]+fib[i-2]; int cnt=0; while(Test--) { N=read(),M=read();num=1; for(int i=1;i<=N;i++) fa[i]=i; for(int i=1;i<=M;i++) { int x=read(),y=read(),z=read(); AddEdge(x,y,z); AddEdge(y,x,z); } int minn=Kruskal(1); for(int i=1;i<=N;i++) fa[i]=i; int maxx=Kruskal(2); bool flag=0; for(int i=1;i<=66;i++) if(minn <= fib[i] && fib[i] <= maxx) {printf("Case #%d: Yes ",++cnt);flag=1;break;} if(flag==0) printf("Case #%d: No ",++cnt); } return 0; }