Gears

Time Limit: 2000ms
Memory Limit: 65536KB

This problem will be judged on ZJU. Original ID: 3789
64-bit integer IO format: %lld      Java class name: Main

 

Bob has N (1 ≤ N ≤ 2*105) gears (numbered from 1 to N). Each gear can rotate clockwise or counterclockwise. Bob thinks that assembling gears is much more exciting than just playing with a single one. Bob wants to put some gears into some groups. In each gear group, each gear has a specific rotation respectively, clockwise or counterclockwise, and as we all know, two gears can link together if and only if their rotations are different. At the beginning, each gear itself is a gear group.

Bob has M (1 ≤ N ≤ 4*105) operations to his gears group:

    • "L u v". Link gear u and gear v. If two gears link together, the gear groups which the two gears come from will also link together, and it makes a new gear group. The two gears will have different rotations. BTW, Bob won't link two gears with the same rotations together, such as linking (a, b), (b, c), and (a, c). Because it would shutdown the rotation of his gears group, he won't do that.
    • "D u". Disconnect the gear u. Bob may feel something wrong about the gear. He will put the gear away, and the gear would become a new gear group itself and may be used again later. And the rest of gears in this group won't be separated apart.
    • "Q u v". Query the rotations between two gears u and v. It could be "Different", the "Same" or "Unknown".
  • "S u". Query the size of the gears, Bob wants to know how many gears there are in the gear group containing the gear u.

Since there are so many gears, Bob needs your help.

Input

Input will consist of multiple test cases. In each case, the first line consists of two integers N and M. Following M lines, each line consists of one of the operations which are described above. Please process to the end of input.

Output

For each query operation, you should output a line consist of the result.

Sample Input

3 7
L 1 2
L 2 3
Q 1 3
Q 2 3
D 2
Q 1 3
Q 2 3
5 10
L 1 2
L 2 3
L 4 5
Q 1 2
Q 1 3
Q 1 4
S 1
D 2
Q 2 3
S 1

Sample Output

Same
Different
Same
Unknown
Different
Same
Unknown
3
Unknown
2

Hint

Link (1, 2), (2, 3), (4, 5), gear 1 and gear 2 have different rotations, and gear 2 and gear 3 have different rotations, so we can know gear 1 and gear 3 have the same rotation, and we didn't link group (1, 2, 3) and group (4, 5), we don't know the situation about gear 1 and gear 4. Gear 1 is in the group (1, 2, 3), which has 3 gears. After putting gear 2 away, it may have a new rotation, and the group becomes (1, 3).

 

Source

Author

FENG, Jingyi
 
解题:并查集
 
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = ;
int fa[maxn],dis[maxn],sum[maxn],mp[maxn],t,n, m;
void init() {
for(int i = ; i <= n+m; i++) {
fa[i] = i;
mp[i] = i;
dis[i] = ;
sum[i] = ;
}
t = n+;
}
int Find(int x) {
if(fa[x] != x) {
int root = Find(fa[x]);
dis[x] += dis[fa[x]];
fa[x] = root;
}
return fa[x];
}
int main() {
char st[];
while(~scanf("%d %d",&n, &m)) {
init();
int x, y;
for(int i = ; i < m; i++) {
scanf("%s",st);
if(st[] == 'L') {
scanf("%d %d",&x, &y);
x = mp[x];
y = mp[y];
int tx = Find(x);
int ty = Find(y);
if(tx != ty) {
sum[tx] += sum[ty];
fa[ty] = tx;
dis[ty] = dis[x]+dis[y]+;
}
} else if(st[] == 'Q') {
scanf("%d %d",&x, &y);
x = mp[x];
y = mp[y];
if(Find(x) != Find(y)) puts("Unknown");
else {
if(abs(dis[x]-dis[y])&) puts("Different");
else puts("Same");
}
} else if(st[] == 'D') {
scanf("%d",&x);
int tx = mp[x];
tx = Find(tx);
sum[tx] -= ;
mp[x] = ++t;
} else if(st[] == 'S') {
scanf("%d",&x);
x = mp[x];
int tx = Find(x);
printf("%d\n",sum[tx]);
}
}
}
return ;
}

xtu read problem training 3 B - Gears的更多相关文章

  1. xtu read problem training 3 A - The Child and Homework

    The Child and Homework Time Limit: 1000ms Memory Limit: 262144KB This problem will be judged on Code ...

  2. xtu read problem training 2 B - In 7-bit

    In 7-bit Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on ZJU. Original ID: 3 ...

  3. xtu read problem training 4 A - Moving Tables

    Moving Tables Time Limit: 2000ms Memory Limit: 65536KB This problem will be judged on ZJU. Original ...

  4. xtu read problem training 4 B - Multiplication Puzzle

    Multiplication Puzzle Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on PKU. O ...

  5. xtu read problem training B - Tour

    B - Tour Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Descriptio ...

  6. xtu read problem training A - Dividing

    A - Dividing Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Descri ...

  7. 2014 Super Training #8 A Gears --并查集

    题意: 有N个齿轮,三种操作1.操作L x y:把齿轮x,y链接,若x,y已经属于某个齿轮组中,则这两组也会合并.2.操作Q x y:询问x,y旋转方向是否相同(等价于齿轮x,y的相对距离的奇偶性). ...

  8. A Gentle Guide to Machine Learning

    A Gentle Guide to Machine Learning Machine Learning is a subfield within Artificial Intelligence tha ...

  9. Bias vs. Variance(1)--diagnosing bias vs. variance

    我们的函数是有high bias problem(underfitting problem)还是 high variance problem(overfitting problem),区分它们很得要, ...

随机推荐

  1. 525 Contiguous Array 连续数组

    给定一个二进制数组, 找到含有相同数量的 0 和 1 的最长连续子数组.示例 1:输入: [0,1]输出: 2说明: [0, 1] 是具有相同数量0和1的最长连续子数组. 示例 2:输入: [0,1, ...

  2. Object类的几个方法

    1.protected Object clone()创建并返回此对象的一个副本. 2. boolean equals(Object obj)指示其他某个对象是否与此对象“相等”. 3. protect ...

  3. [转]写给Git初学者的7个建议

    本文转自:http://www.open-open.com/news/view/b7227e 阅读目录 第一条:花时间去学习 Git 的基本操作 第二条:从简单的 Git 工作流开始 第四条:理解分支 ...

  4. [转]EntityFramework之领域驱动设计实践

    本文转自:http://www.cnblogs.com/daxnet/archive/2010/11/02/1867392.html Entity Framework之领域驱动设计实践 EntityF ...

  5. [转].NET 4 并行(多核)编程系列之二 从Task开始

    本文转自:http://www.cnblogs.com/yanyangtian/archive/2010/05/22/1741379.html .NET 4 并行(多核)编程系列之二 从Task开始 ...

  6. hihocoder1779 公路收费

    思路: 枚举每个点做根即可. 实现: #include <bits/stdc++.h> using namespace std; typedef long long ll; const l ...

  7. java 获取ip地址

    1.使用WIFI 首先设置用户权限 <uses-permission android:name="android.permission.ACCESS_WIFI_STATE"& ...

  8. Knockout-了解Observable与computed

    KO是什么? KO不是万能的,它的出现主要是为了方便的解决下面的问题: UI元素较多,用户交互比较频繁,需要编写大量的手工代码维护UI元素的状态.样式等属性? UI元素之间关系比较紧密,比如操作一个元 ...

  9. ubuntu服务器切换语言

    如果在安装Ubuntu Server时选择了中文,在系统安装完毕后,默认是中文,在操作时经常会显示乱码,如果需要设置回英文,则修改/etc/default/locale,将 LANG="cn ...

  10. H3C S5024P交换机 vlan实验

    H3C S5024P交换机第二次vlan实验 实验1 与交换机端口G0/1和G0/2相连的PC1与PC2属于VLAN 1,与G0/3和G0/4相连的PC3和PC4属于VLAN 2,PC1.PC2.PC ...