HDU 4786 生成树 并查集+极大极小值 黑白边 确定选择白边的数量
题意:
给定一个无向图
n 个点 m条无向边
u v val
val == 1 表示边(u, v) 为白边
问能否找到n个点的生成树, 使得白边数为斐波那契数
思路:
并查集求图是否连通( 是否存在生成树)
求出 最多白边树 的 白边数量
求出 最少白边树 的 白边数量
若[最少, 最多] 区间内存在斐波那契数 ,则满足条件
(也就是说,白边的数量是连续有解的)
//YY得证
#include<iostream> #include<stdio.h> #include<algorithm> #include<string> #include<queue> #include<string.h> #include<map> #include<set> #include<stack> #include<vector> #include<math.h> #include<algorithm> #define N 101010 #define inf 10000000 using namespace std; inline int Min(int a,int b){return a>b?b:a;} inline int Max(int a,int b){return a<b?b:a;} int f[N]; int find(int x){return x==f[x]?x:(f[x] = find(f[x]));} void Union(int u, int v){ int fu = find(u), fv = find(v); if(fu>fv) f[fu] = fv; else f[fv] = fu; } set<int>fib; int n, m; struct node{ int u,v,c; }edge[N]; int edgenum; bool cmp1(node a,node b){return a.c<b.c;} bool cmp2(node a,node b){return a.c>b.c;} int main(){ int T, Cas = 1;scanf("%d",&T); int i, j, col; fib.clear(); fib.insert(1); fib.insert(2); j=1; for(i=2;i<=N;){ fib.insert(i+j); int lala = i; i = i+j; j = lala; } while(T--){ scanf("%d %d", &n, &m); for(i=1;i<=n;i++)f[i] = i; edgenum = 0; while(m--){ int u,v; scanf("%d %d %d",&u,&v,&col); edge[edgenum].u = u; edge[edgenum].v = v; edge[edgenum++].c = col; int fx = find(u), fy = find(v); if(fx == fy)continue; Union(fx,fy); } printf("Case #%d: ",Cas++); for(i=1;i<=n;i++)find(i); bool su = true; for(i=1;i<=n;i++) if(f[i]!=f[1]) {su = false; break;} if(su == false) {printf("No\n");continue;} for(i=1;i<=n;i++)f[i] = i; sort(edge, edge+edgenum, cmp1); int size = 0, bl=0, bm=0; for(i=0;i<edgenum;i++) { int u =edge[i].u, v=edge[i].v; int fu = find(u), fv=find(v); if(fu == fv)continue; size++; bl+= edge[i].c; Union(fu, fv); if(size==n-1)break; } for(i=1;i<=n;i++)f[i] = i; sort(edge, edge+edgenum, cmp2); size = 0; for(i=0;i<edgenum;i++) { int u =edge[i].u, v=edge[i].v; int fu = find(u), fv=find(v); if(fu == fv)continue; size++; bm+= edge[i].c; Union(fu, fv); if(size==n-1)break; } if(fib.upper_bound(bl) == fib.end() ){printf("No\n");continue;} if((*fib.lower_bound(bl) )>bm){printf("No\n");continue;} printf("Yes\n"); } return 0; } /* 4 2 1 2 1 3 4 1 */
posted on 2013-11-19 11:26 love so much 阅读(261) 评论(0) 编辑 收藏 举报