Problem W

Posted 2023-02-17 tansanity

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>

简单题意：   有编号为1-N的人在体育馆里，输入为 A B X ,代表二者之间的距离为X。然后给出一组数据，求出有多少错误的数据，即有冲突的数据，当前的关系与之前的关系有冲突。 解题思路形成过程：   运用并查集求出带权的树，这是核心的思想。并查集的代码模板可以完全胜任这个程序，所以问题也就自然而然的解决。关键是并查集的运用。 感想：   图论这个专题，其实新颖的东西并不多，只要掌握了核心，一招鲜吃遍天下。 AC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;