题目：
Zjnu Stadium
**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3280 Accepted Submission(s): 1267
**
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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.\
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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.
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Output
For every case:
Output R, represents the number of incorrect request.\
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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
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Sample Output
2
Hint
Hint:
（PS： the 5th and 10th requests are incorrect）
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Source
2009 Multi-University Training Contest 14 - Host by ZJNU
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Statistic | Submit | Discuss | Note
思路：
给出一个n,m,n表示的是有n 个人，m表示的是 有m 对关系： 接下来输入的就是这m对关系，a,b,x；表示的是a，b相距x个距离；然后判断输入的是否与这个数的上面的数信息一致， 输出不一致的数目,出现错误的可能性有一种，就是两个人到根节点的距离相同，这样就出现矛盾了，所以用dis[]数组来记录当前点到根节点的距离，然后dis[fy]=dis[x]+s-dis[y],这一句代码的推导类似于向量，如下图：\
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根据向量的减法，就可以推出这个公式
代码：
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define N 50000+20
#define M 200010
#define MOD 1000000000+7
#define inf 0x3f3f3f3f
using namespace std;
int pre[N], dis[N],cnt;//dis表示这个节点到根节点的距离
void init(int n)
{
for(int i=1; i<=n; i++)
{
pre[i]=i;
dis[i]=0;
}
}
int find(int x)
{
if(pre[x]==x)
return x;
int temp=pre[x];//找到x的父亲节点
pre[x]=find(pre[x]);//找到x的根节点
dis[x]+=dis[temp];//更新x到根节点的距离
return pre[x];
}
void mix(int x,int y,int s)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
{
pre[fy]=fx;
dis[fy]=dis[x]+s-dis[y];//类似于向量
}
else
{
if(dis[y]-dis[x]!=s)//如果y到根节点的距离减去x到根节点的距离不是s，那么就证明这一句错误
cnt++;
}
}
int main()
{
int n,m,a,b,c;
while(~scanf("%d%d",&n,&m))
{
cnt=0;
init(n);
for(int i=1; i<=m; i++)
{
scanf("%d%d%d",&a,&b,&c);
mix(a,b,c);
}
printf("%d\n",cnt);
}
return 0;
}
PS：借一下这道题来理解理解带权并查集
