Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5934 Accepted Submission(s): 1845 Problem Descrip ...
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5934 Accepted Submission(s):
1845
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
Recommend We have carefully selected several similar problems for you: 6263 6262 6261 6260 6259 和昨天ysy講的那道題差不多 而且這道題在題目中直接給提示了——》黑邊為0，白邊為1 這樣的話我們做一個最小生成樹和一個最大生成樹 如果在這兩個值的範圍內有斐波那契數，就說明滿足條件 簡單證明：
對於最小生成樹來說，任意刪除一條邊，並加入一條沒有出現過的邊，這樣的話權值至多加1，邊界為最大生成樹
#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\n",++cnt);flag=1;break;} if(flag==0) printf("Case #%d: No\n",++cnt); } return 0; }
