CodeForces 639B Bear and Forgotten Tree 3(水题)
来源:互联网 发布:软件著作权许可使用 编辑:程序博客网 时间:2024/05/21 18:41
A tree is a connected undirected graph consisting of n vertices and n - 1 edges. Vertices are numbered 1 through n.
Limak is a little polar bear and Radewoosh is his evil enemy. Limak once had a tree but Radewoosh stolen it. Bear is very sad now because he doesn't remember much about the tree — he can tell you only three values n, d and h:
- The tree had exactly n vertices.
- The tree had diameter d. In other words, d was the biggest distance between two vertices.
- Limak also remembers that he once rooted the tree in vertex 1 and after that its height was h. In other words, h was the biggest distance between vertex 1 and some other vertex.
The distance between two vertices of the tree is the number of edges on the simple path between them.
Help Limak to restore his tree. Check whether there exists a tree satisfying the given conditions. Find any such tree and print its edges in any order. It's also possible that Limak made a mistake and there is no suitable tree – in this case print "-1".
The first line contains three integers n, d and h (2 ≤ n ≤ 100 000, 1 ≤ h ≤ d ≤ n - 1) — the number of vertices, diameter, and height after rooting in vertex 1, respectively.
If there is no tree matching what Limak remembers, print the only line with "-1" (without the quotes).
Otherwise, describe any tree matching Limak's description. Print n - 1 lines, each with two space-separated integers – indices of vertices connected by an edge. If there are many valid trees, print any of them. You can print edges in any order.
5 3 2
1 21 33 43 5
8 5 2
-1
8 4 2
4 85 72 38 12 15 61 5
Below you can see trees printed to the output in the first sample and the third sample.
#include<iostream>#include<stack>#include<string.h>#include<stdio.h>#include<algorithm>#include<queue>using namespace std;const int mod=1e9+7;const int N=10005;int a[N],n,m,d,h;int main(){ int i,j,k; while(scanf("%d%d%d",&n,&d,&h)!=EOF) { if(d>2*h) { printf("-1\n"); continue; } if(d==h) { if(d==1) { if(n==2) printf("1 2\n"); else printf("-1\n"); continue; } if(d==n-1) { for(i=1; i<n; i++) printf("%d %d\n",i,i+1); continue; } else { for(i=1; i<=h; i++) { printf("%d %d\n",i,i+1); } i++; while(i<=n) printf("2 %d\n",i++); } continue; } int x=d-h; for(i=1; i<=h; i++) { printf("%d %d\n",i,i+1); } i++; while(i<=n) { printf("1 %d\n",i); i++; for(j=1; j<x; j++) { if(i>n) break; printf("%d %d\n",i-1,i); i++; } } } return 0;}
- CodeForces 639B Bear and Forgotten Tree 3(水题)
- CodeForces 639 B.Bear and Forgotten Tree 3(构造)
- Codeforces 639B Bear and Forgotten Tree 3 【构造】
- [杂题 图论 构造] Codeforces #639B. Bear and Forgotten Tree 3
- Code Forces Bear and Forgotten Tree 3 639B
- CodeForces 658C Bear and Forgotten Tree 3(构造)
- Codeforces 658C Bear and Forgotten Tree 3【构造】
- CodeForces-657A-Bear and Forgotten Tree 3
- CodeForces 657A Bear and Forgotten Tree 3
- Codeforces 658C Bear and Forgotten Tree 3【思维】
- Codeforces--658C--Bear and Forgotten Tree 3(模拟&&技巧)(好题)
- Codeforces 653E:Bear and Forgotten Tree 2
- codeforces 653E. Bear and Forgotten Tree 2 连通问题
- CodeForces 653E Bear and Forgotten Tree 2
- codeforces 658C Bear and Forgotten Tree 3构造-多么痛的领悟
- Codeforces 658 C. Bear and Forgotten Tree 3(树的构造)
- VK Cup 2016 - Round 1 B. Bear and Forgotten Tree 3
- codeforce 658C Bear and Forgotten Tree 3
- MySQL索引原理及慢查询优化
- 详解网络传输中的三张表,MAC地址表、ARP缓存表以及路由表
- 20160330 HDU5651 xiaoxin juju needs help(组合数公式)
- 重新安装包
- sql中的group by 和 having 用法解析
- CodeForces 639B Bear and Forgotten Tree 3(水题)
- lightoj 1173 - The Vindictive Coach 计数类DP
- quartz的两个点
- 浅析java反射机制
- sata硬盘
- sleep wait区别
- Selenium源码学习
- 蓝桥杯_算法提高_新建MicrosoftWord文档
- bzoj 2821: 作诗(Poetize)