训练4 习题23

代码：

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.<br>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).<br>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.<br>

 

Input

There are many test cases:<br>For every case: <br>The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.<br>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.<br><br>

 

Output

For every case: <br>Output R, represents the number of incorrect request.<br>

 

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
<div style='font-family:Times New Roman;font-size:14px;background-color:F4FBFF;border:#B7CBFF 1px dashed;padding:6px'><div style='font-family:Arial;font-weight:bold;color:#7CA9ED;border-bottom:#B7CBFF 1px dashed'><i>Hint</i></div>
Hint:
（PS： the 5th and 10th requests are incorrect）
</div>

 

思路：

用带权并查集做     合并这a1,a2两棵树   a1,a2之间的相对距离，可以得到rank[root_a] = rank[a1]+x-rank[a2]

代码：

#include<cstdio>  
#include<cmath>  
using namespace std;  
#define N 50005  
int f[N], rank[N], n, m;  
  
void init(){  
    for(int i=0; i<=n; ++i)  
        f[i]=i, rank[i]=0;  
}  
int find(int x){  
    if(x==f[x]) return f[x];  
    int t=f[x];  
    f[x] = find(f[x]);  
    rank[x] += rank[t];  
    return f[x];  
}  
bool Union(int x,int y, int m){  
    int a=find(x), b=find(y);  
    if(a==b){  
        if(rank[x]+m!=rank[y])  
            return false;  
        return true;  
    }  
    f[b] = a;  
    rank[b] = rank[x]+m-rank[y];  
    return true;  
}  
  
int main(){  
    int a,b,x;  
    while(~scanf("%d%d",&n,&m)){  
        init();  
        int cnt=0;  
        for(int i=0; i<m; ++i){  
            scanf("%d%d%d",&a,&b,&x);  
            if(!Union(a, b, x)){  
                ++cnt;  
            }  
        }  
        printf("%d\n",cnt);  
    }  
    return 0;  
}