HDU 3407.Zjnu Stadium 加权并查集

Posted 2020-09-26 GeekZRF

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3726    Accepted Submission(s): 1415

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）

 

 

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=3047

题意：n个点，m个关系。每个关系有3个整数a，b，x表示为a到b的距离为x。问m个关系中有几个不合理的。

思路：并查集，在压缩路径的时候顺便处理一下距离dist[x]+=dist[pa[x]]。关系合并时距离关系为：dist[fb]=dist[a]+x-dist[b]（fb为b的祖先；dist[a],diat[b]分别表示a，b到祖先的距离）。

 

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
#define PI acos(-1.0)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=1e5+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e13+7;
int pa[maxn];
int dist[maxn];
void init(int n)
{
    for(int i=1; i<=n; i++)
        pa[i]=i,dist[i]=0;
}
int findset(int x)
{
    if(pa[x]==x) return x;
    int fax=pa[x];
    pa[x]=findset(pa[x]);
    dist[x]+=dist[fax];
    return pa[x];
}
void unit(int x,int y,int w)
{
    int fx=findset(x),fy=findset(y);
    pa[fx]=fy;
    dist[fx]=dist[y]+w-dist[x];
}
bool same(int x,int y,int w)
{
    if(findset(x)!=findset(y)) return false;
    else if(dist[findset(x)]==dist[y]+w-dist[x]) return false;
    else return true;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int ans=0;
        init(n);
        for(int i=1; i<=m; i++)
        {
            int a,b,x;
            scanf("%d%d%d",&a,&b,&x);
            if(same(a,b,x)) ans++;
            else unit(a,b,x);
        }
        cout<<ans<<endl;
    }
    return 0;
}

加权并查集