HDU5631 Rikka with Graph
来源:互联网 发布:最基本的网络拓扑结构 编辑:程序博客网 时间:2024/05/29 14:17
Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one) and delete them from the graph.
Yuta wants to know the number of the ways to choose the edges in order to make the remaining graph connected.
It is too difficult for Rikka. Can you help her?
For each testcase, the first line contains a number
Then n+1 lines follow. Each line contains two numbers
131 22 33 11 3
9
具有n个顶点和n+1个边的非连通图形。可以选择一些边(至少一个),并从图中删除它们。想得出选择边缘的方法的数量,以使剩余的图形连通。
输入:
line 1:T个样例
line 2:N个顶点
line 3:N+1行边
输出:
选择删除边数的方法数量。
解题思路:
1.建立一个结构体来存储这N个顶点
struct number{
int x,y;
int flag;
}a[107];
int s[107]; //来存储父节点
int num[107]; ///来存储秩
2.因为N个顶点的话,最少必须有N-1条边进行连通。所以两个循环来进行选择删除边数。
for(i=0;i<=n;i++)
{
a[i].flag=0;
//printf("删除第%d条边!\n",i+1);
judge();
for(j=i+1;j<=n;j++)
{
a[j].flag=0;
//printf("删除第%d条边!\n",j+1);
judge();
//printf("恢复第%d条边!\n",j+1);
a[j].flag=1;
}
a[i].flag=1;
//printf("恢复第%d条边!\n",i+1);
}
3.判断删除两条边数后,是否是一个连通的图形。
num=0;
for(j=1;j<=n;j++)
{
if(s[j]==j)
num+=1;
}
if(num==1)
{
//printf("找到一种方案!\n");
number+=1;
}
也就是说,在父节点数组中,只有一个节点的父节点是它本身时,其余节点都是该节点的子节点,该方案就满足条件。
坑点总结:
1.只要头脑清醒,该题不是问题。
程序代码:
#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std; struct number{int x,y;int flag;}a[107];int s[107]; int num[107];int n,x,x1;int number;int find(int i){while(i!=s[i])i=s[i];return i;}void judge(){int i,j,p,q;int count;for(i=1;i<=n;i++)s[i]=i; for(i=0;i<=n;i++){if(a[i].flag==1){ p=find(a[i].x); q=find(a[i].y); if(q!=p) s[p]=q; //printf("%d的父节点是%d\n",p,q); }}count=0;for(j=1;j<=n;j++){if(s[j]==j)count+=1;if(count>1)break;}if(count==1)//printf("找到一种方案!\n"); number++;}int main(){int t,i,j;scanf("%d",&t);while(t--){number=0;scanf("%d",&n);for(i=0;i<=n;i++){scanf("%d%d",&x,&x1);a[i].x=x;a[i].y=x1;a[i].flag=1;}for(i=0;i<=n;i++){a[i].flag=0;//printf("删除第%d条边!\n",i+1); judge();for(j=i+1;j<=n;j++){a[j].flag=0;//printf("删除第%d条边!\n",j+1); judge();//printf("恢复第%d条边!\n",j+1);a[j].flag=1;}a[i].flag=1;//printf("恢复第%d条边!\n",i+1);}printf("%d\n",number); } return 0;
}
- HDU5631 Rikka with Graph
- Rikka with Graph||HDU5631
- HDU5631 Rikka with Graph
- HDU5631--Rikka with Graph--并查集
- HDU5631——Rikka with Graph
- HDU5631——Rikka with Graph【并查集】
- Rikka with Graph
- Bc Rikka with Graph
- Rikka with Graph<hdoj5631>
- Rikka with Graph HDU
- Rikka with Graph
- HDU6090 Rikka with Graph
- HDU6090 Rikka with Graph
- HDU6090 Rikka with Graph
- hdu6090 Rikka with Graph
- Rikka with Graph
- Rikka with Graph
- hdu6090 Rikka with Graph
- TOAD连接Oracle数据库失败:OCI_INVALID_HANDLE解决
- 第一个jsp
- 初始化项目——快捷键
- 二叉树的递归和非递归前、中、后序遍历
- c++的浅复制与深复制
- HDU5631 Rikka with Graph
- 剑指offer——合并链表
- set与单词数
- Json,List,Map的数据格式详解。
- 控制并发线程数的Semaphore和线程之间的数据交换Exchanger
- 给定一个字符串,问是否能够通过添加一个字母将其变成“回文串”
- int转换成NSData
- centos7 yum安装配置redis 并设置密码
- Git远程连接Github常见问题