hdu3047 Zjnu Stadium （并查集）

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1686    Accepted Submission(s): 634

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）

 

题意：懒得说了。。比赛的时候一直看错题意我擦。。

思路： 用并查集维护相对距离，若出现一个操作，两人的祖先相同，那么可以根据这个算出这两人的相对距离，若与操作所给的距离不一样，则为错误操作。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <utility>
#include <queue>
#include <stack>
using namespace std;
const int INF=1<<30;
const double eps=1e-6;
const int N = 50005;
const int mod = 300;

int fa[N],d[N];
int n,m;

void init()
{
    for(int i=1;i<=n;++i)
        fa[i]=i, d[i]=0;
}

int find(int x)
{
    //return fa[x]==x? x: fa[x]=find(fa[x]);
    if(fa[x]==x)
        return x;
    int u=find(fa[x]);
    d[x]+=d[fa[x]];
    fa[x] = u;
    return u;
}

bool equ(int a,int b)
{
    return (b-a)%mod == 0;
}

void run()
{
    init();
    int a,b,x;
    int ta,tb;
    int ans=0;
    while(m--)
    {
        scanf("%d%d%d",&a,&b,&x);
        ta = find(a);
        tb = find(b);
        if(ta!=tb)
        {
            fa[tb]=ta;
            d[tb]=d[a]-d[b]+x;
        }
        else
        {
            if(d[b]-d[a]!=x)
                ++ans;
        }
    }
    printf("%d
",ans);
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
        run();
    return 0;
}