Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2672    Accepted Submission(s): 1028

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

题意：就是一个无限大的场地里面坐了很多人,每两个人之间都有一个距离。如果后面的描述和前面的冲突，那么则认为这条是错误的。问总共有多少条错误的描述。

题解：向量偏移的并查集。详细了解可以参考我的博客：http://www.cnblogs.com/liyinggang/p/5327055.html

我们设某个点到根节点的距离a->root为sum[a]，然后套用公式就OK

当描述的两个人已经位于同一棵子树了

a->b = a->root - b->root

还没有位于同一棵子树时

roota ->rootb = b->rootb + a->b - a->roota

方向千万不要搞错。

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int N =50001;
int father[N];
int sum[N];  ///记录当前结点到根结点的距离

int _find(int x){
    if(x!=father[x]){
        int t = father[x];
        father[x] = _find(father[x]);
        sum[x]+=sum[t];
    }
    return father[x];
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=0;i<=n;i++){
            father[i] = i;
            sum[i] = 0;
        }
        int ans = 0;
        while(m--){
            int a,b,v;
            scanf("%d%d%d",&a,&b,&v);
            int roota = _find(a);
            int rootb = _find(b);
            if(roota==rootb){
                if(sum[a]-sum[b]!=v) ans++;
            }
            else{
                father[roota] = rootb;
                sum[roota] = -sum[a]+sum[b]+v;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}