Hdu 2047 Zjnu Stadium(带权并查集)

Zjnu Stadium
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3179 Accepted Submission(s): 1224
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

/*
带权并查集.
find逐层更新cnt值. 
用向量搞出merge关系.
cnt[i]表示从i的祖宗到i的距离. 
有cnt[l2]=z+cnt[x]-cnt[y].
*/
#include<iostream>
#include<cstdio>
#define MAXN 100001
using namespace std;
int father[MAXN],cnt[MAXN],n,m,ans;
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
    return x*f;
}
int find(int x)
{
    if(x==father[x]) return x;
    int fa=father[x];
    father[x]=find(father[x]);
    cnt[x]+=cnt[fa];
    return father[x];
}
void merge(int x,int y,int z)
{
    int l1=find(x),l2=find(y);
    father[l2]=l1;
    cnt[l2]=z+cnt[x]-cnt[y];
    return ;
}
int main()
{
    int x,y,z;
    while(~scanf("%d%d",&n,&m))
    {
        ans=0;
        for(int i=1;i<=n;i++) father[i]=i,cnt[i]=0;
        while(m--)
        {
            x=read(),y=read(),z=read();
            int l1=find(x),l2=find(y);
            if(l1==l2){if(cnt[y]-cnt[x]!=z) ans++;}
            else merge(x,y,z);
        }
        printf("%d\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/nancheng58/p/10068128.html