uva1160 - X-Plosives (并查集)
来源:互联网 发布:农村污水治理数据 编辑:程序博客网 时间:2024/05/22 12:22
X-Plosives
A secret service developed a new kind of explosive that attain its volatile property only when a specific association of products occurs. Each product is a mix of two different simple compounds, to which we call a
You are not a secret agent but only a guy in a delivery agency with one dangerous problem: receive binding pairs in sequential order and place them in a cargo ship. However, you must avoid placing in the same room an explosive association. So, after placing a set of pairs, if you receive one pair that might produce an explosion with some of the pairs already in stock, you must refuse it, otherwise, you must accept it.
An example. Let’s assume you receive the following sequence: A+B, G+B, D+F, A+E, E+G, F+H. You would accept the first four pairs but then refuse E+G since it would be possible to make the following explosive with the previous pairs: A+B, G+B, A+E, E+G (4 pairs with 4 simple compounds). Finally, you would accept the last pair, F+H.
Compute the number of refusals given a sequence of binding pairs.
Input
The input will contain several test cases, each of them as described below. Consecutive test cases are separated by a single blank line.
Instead of letters we will use integers to represent compounds. The input contains several lines. Each line (except the last) consists of two integers (each integer lies between 0 and 105) separated by a single space, representing a binding pair. The input ends in a line with the number –1. You may assume that no repeated binding pairs appears in the input.
Output
For each test case, a single line with the number of refusals.
Sample Input
1 2
3 4
3 5
3 1
2 3
4 1
2 6
6 5
-1
Sample Output
3
#include <iostream>
#include <stdio.h>
using namespace std;
const int maxn=100000+10;
int p[maxn];
int find(int x){return p[x]!=x?p[x]=find(p[x]):x;}
int main()
{int x,y,i;
while(scanf("%d",&x)==1)
{for(i=0;i<maxn;i++)p[i]=i;
int re=0;
while(x!=-1)
{cin>>y;
x=find(x);y=find(y);
if(x==y)re++;
else p[x]=y;
cin>>x;
}
cout<<re<<endl;
}
return 0;
}
- uva1160 - X-Plosives (并查集)
- 并查集uva1160 - X-Plosives
- BNU X-Plosives(并查集) @
- uva1160 X-Plosives
- 并查集uva1160
- UVALive(LA) 3644 X-Plosives (并查集)
- uva 1160 - X-Plosives(并查集)
- uva 1160 - X-Plosives(并查集)
- ACM LA3644: X-Plosives(并查集)
- UVA 1160 X-Plosives(并查集)
- UVALive - 3644X-Plosives(并查集)
- UVALive - 3644 - X-Plosives (并查集!!)
- UVA 1160 - X-Plosives(并查集)
- uvalive 3644 X-Plosives(并查集)
- LA3644 X-Plosives(无向图中找环,并查集)
- UVALive 3644 X-Plosives (并查集)
- LA 3644 - X-Plosives,并查集
- LA 3644 X-Plosives / 并查集
- MyEclipse10 安装JBPM5.4插件最简单方法
- The maximum-subarray problem
- hdu 2987 邂逅明下
- java 图像旋转
- 谈MongoDB的应用场景
- uva1160 - X-Plosives (并查集)
- uva 10458 - Cricket Ranking(容斥+高精度)
- TOJ4646划分数
- 关于HRESULT句柄的说明
- spring AOP代理模式(SSH学习第8天)
- 算法分析-交换排序(冒泡排序 & 快速排序)
- hdu3836 强联通水题
- UVa 10534 - Wavio Sequence (简单DP 最长上升下降子序列)
- 索引 索引 索引