HDU3047带权并查集

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2419    Accepted Submission(s): 922

Problem Description

In 12th Zhejiang College Students Games 2007, there was a new stadium built in Zhejiang Normal University. It was a modern stadium which could hold thousands of people. The audience Seats made a circle. The total number of columns were 300 numbered 1--300, counted clockwise, we assume the number of rows were infinite.
These days, Busoniya want to hold a large-scale theatrical performance in this stadium. There will be N people go there numbered 1--N. Busoniya has Reserved several seats. To make it funny, he makes M requests for these seats: A B X, which means people numbered B must seat clockwise X distance from people numbered A. For example: A is in column 4th and X is 2, then B must in column 6th (6=4+2).
Now your task is to judge weather the request is correct or not. The rule of your judgement is easy: when a new request has conflicts against the foregoing ones then we define it as incorrect, otherwise it is correct. Please find out all the incorrect requests and count them as R.

 

Input

There are many test cases:
For every case: 
The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.
Then M lines follow, each line has 3 integer A(1<=A<=N), B(1<=B<=N), X(0<=X<300) (A!=B), separated by a space.

 

Output

For every case: 
Output R, represents the number of incorrect request.

 

Sample Input

   
   
   
   

    
    
    
    10 10
1 2 150
3 4 200
1 5 270
2 6 200
6 5 80
4 7 150
8 9 100
4 8 50
1 7 100
9 2 100
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2

    
    
    
    
 
     
 
      Hint 
     
 Hint: （PS： the 5th and 10th requests are incorrect） 
    
    
    
    
     
   
   
   
   

 

题目大意：

       魔术师为了表演更加有趣，就是让大家一直找座位做，比如说A B 2 就是B坐在A的右边两个位置。产生矛盾的条件就是两个人坐在同一个位置上了。

遇到的问题和思路：

        刚看到这道题看不懂，以为怎么是圈，然后看了一下题解，才明白。。。这道题题目貌似没有说明白。

        这道题目的意思就是，所有人都在一条直线上坐着（请忽视圈）。然后用并查集来储存。

        这道题目的最难点就是如何处理根。通过画图可以获得，刚开始假设 A 和 B 的距离是100，他们是一棵树；C和D的距离是200，他们也是一棵树。然后这个时候B和D的距离是400，那么就说明A与B所在的这棵树和C与D所在的这棵树连在一起了。假设我们把根定做是A，那么C距离A的位置就是400 - 200 + 100 = 100，然后通过数学归纳法得到，两棵树合并，被合并的那棵树的根到与之合并的根的距离为：B和D的距离 - CD + AB就好了

给出AC代码：

#include<cstdio> #include<cstring> using namespace std; const int MAX_N = 50000 + 50; int n, m; int par[MAX_N], level[MAX_N]; void init(){ for(int i = 1; i <= n; i++){ par[i] = i; level[i] = 0; } } int find(int x){ if(par[x] == x)return x; int t = par[x]; //level[x] += level[t];//不知道为啥，放在这里就是WA par[x] = find(par[x]); level[x] += level[t];//每一次寻找根的时候都会加一次。 return par[x];//路径压缩 } bool same(int x, int y){ return find(x) == find(y); } void unite(int x, int y, int dis){ int a = find(x); int b = find(y); par[b] = a; level[b] = dis - level[y] + level[x]; } int main(){ while(scanf("%d%d", &n, &m) != EOF){ init(); int a, b, c, cnt = 0; for(int i = 1; i <= m; i++){ scanf("%d%d%d", &a, &b, &c); if(!same(a, b)){ unite(a, b, c); } else { if(level[a] + c != level[b])cnt++; } } printf("%d\n", cnt); } return 0; } </cstring></cstdio>