带权并查集
例：POJ1182食物链
动物王国中有三类动物A,B,C，这三类动物的食物链构成了有趣的环形。A吃B， B吃C，C吃A。
现有N个动物，以1－N编号。每个动物都是A,B,C中的一种，但是我们并不知道它到底是哪一种。
有人用两种说法对这N个动物所构成的食物链关系进行描述：
第一种说法是"1 X Y"，表示X和Y是同类。
第二种说法是"2 X Y"，表示X吃Y。
此人对N个动物，用上述两种说法，一句接一句地说出K句话，这K句话有的是真的，有的是假的。当一句话满足下列三条之一时，这句话就是假话，否则就是真话。
1） 当前的话与前面的某些真的话冲突，就是假话；
2） 当前的话中X或Y比N大，就是假话；
3） 当前的话表示X吃X，就是假话。
你的任务是根据给定的N（1 <= N <= 50,000）和K句话（0 <= K <= 100,000），输出假话的总数。
Input
第一行是两个整数N和K，以一个空格分隔。
以下K行每行是三个正整数 D，X，Y，两数之间用一个空格隔开，其中D表示说法的种类。
若D=1，则表示X和Y是同类。
若D=2，则表示X吃Y。
Output
只有一个整数，表示假话的数目。
Sample Input
100 7
1 101 1
2 1 2
2 2 3
2 3 3
1 1 3
2 3 1
1 5 5
Sample Output
3详细注解：https://blog.youkuaiyun.com/qq_29660153/article/details/68076314
相似题目：Bug's Life hdu-1829
Problem Description
Background
Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs.
Problem
Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it.
Input
The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one.
Output
The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong.
Sample Input
2
3 3
1 2
2 3
1 3
4 2
1 2
3 4
Sample Output
Scenario #1:
Suspicious bugs found!
Scenario #2:
No suspicious bugs found!
Hint
Huge input,scanf is recommended.
题意大概是随便给你一堆虫子，拿出来两个交配要是两个为同性就是出bug了！
注意输入输出！
当前点与根节点的关系可以看成向量
如图：
当ra != rb 时进行虚线的变化， srb指b与rb之间的关系，sra同srb
具体代码如下：
#include <iostream>
#include <cstdio>
using namespace std;
int root[2005], sroot[2005];
int rfind(int a, int &ans)
{
ans = 0;
while(root[a] != a)
{
ans = (ans + sroot[a]) % 2;
a = root[a];
}
return a;
}
int main()
{
int cn, T;
scanf("%d", &T);
for(cn = 1; cn <= T; cn++)
{
int m, n, i, ans = 0;
scanf("%d%d",&m, &n);
for(i = 0; i <= m; i++)
{
root[i] = i;
sroot[i] = 0;
}
for(i = 0; i < n; i++)
{
int a, b;
scanf("%d%d",&a, &b);
int ra, rb, sra, srb;
ra = rfind(a, sra);
rb = rfind(b, srb);
if(a > m || a < 1 || b > m || b < 1)
{
ans++;
continue;
}
if(ans == 0)
{
if(ra != rb)
{
root[rb] = ra;
sroot[rb] = (sra - srb + 1) % 2;
}
else
{
if(sra == srb)
{
ans++;
}
}
}
}
if(cn != 1)
cout << endl;
cout << "Scenario #" << cn << ":" << endl;
if(ans != 0)
{
cout << "Suspicious bugs found!" << endl;
}
else
{
cout << "No suspicious bugs found!" << endl;
}
}
return 0;
}
另一道带权并查集的题，或者说另一种思想。。。。。。
poj：1988
Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 28794 | Accepted: 10105 | |
| Case Time Limit: 1000MS | ||
Description
Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:
moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.
Write a program that can verify the results of the game.
Input
* Line 1: A single integer, P
* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.
Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.
Output
Print the output from each of the count operations in the same order as the input file.
Sample Input
6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4
Sample Output
1 0 2
大致意思就是有一堆箱子，有两种操作，一种是把a以及a下面的箱子放到b上，另一种就是查询b下面有多少箱子！
这道题借鉴了网上的一些代码，其中让我觉得巧妙的是，root【i】数组被赋值为-1，每一次合并就将root【a】+root【b】这样就得到了a下面有-n（注意是负的！）个箱子，最后结果用根减去当前的在减一就是答案了。
代码如下：
#include <iostream>
#include <cstdio>
using namespace std;
int root[30005], sroot[30005];
int getroot(int a)
{
if(root[a] < 0)
{
return a;
}
int p = root[a];
root[a] = getroot(p);
sroot[a] += sroot[p];
return root[a];
}
int main()
{
int p, i;
for(i = 0; i < 30005; i++)
{
root[i] = -1;
sroot[i] = 0;
}
cin >> p;
while(p--)
{
int a, b;
char c;
getchar();
scanf("%c",&c);
if(c == 'C')
{
scanf("%d", &a);
int ra = getroot(a), ans;
ans = (-root[ra]) - sroot[a] - 1;
cout << ans << endl;
}else if(c == 'M')
{
scanf("%d%d", &a, &b);
int ra, rb;
rb = getroot(b);
ra = getroot(a);
if(ra != rb)
{
sroot[rb] = -root[ra];
root[ra] += root[rb];
root[rb] = ra;
}
}
}
return 0;
}
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