HDU 3047 Zjnu Stadium

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4273    Accepted Submission(s): 1627

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）

 

思路：

带权并查集。。感觉全部都是用向量来确定权值之间的关系。。之前向量方向弄反了一条，找了几个小时才找到错，心态爆炸。。。

 

实现代码：

#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long
const int M = 100009;
ll pre[M],rel[M];
void init(int n){
    for(int i=0;i<=n;i++){
        pre[i] = i;
        rel[i] = 0;
    }
    return ;
}

int find(int x){
    if(x==pre[x])
        return x;
    int temp = pre[x];
    pre[x] = find(temp);
    rel[x] = rel[x] + rel[temp];
    return pre[x];
}

int mix(int x,int y,int d){
    int fx = find(x);
    int fy = find(y);
    if(rel[y] - rel[x] != d&&fx==fy)
    return 1;
    pre[fy] = fx;
    rel[fy] = rel[x] + d - rel[y];
    return 0;
}

int main(){
    int n,m,x1,y1,d1,ans;
    while(scanf("%d%d",&n,&m)!=EOF){
        ans = 0;
        init(n);
        while(m--){
            scanf("%d%d%d",&x1,&y1,&d1);
            ans += mix(x1,y1,d1);
        }
        printf("%d
",ans);
    }
    return 0;
}