Zjnu Stadium_牛客博客

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

Sample Output

Hint

Hint:
（PS： the 5th and 10th requests are incorrect）

C++版本一

题目大意:
有n个人坐在zjnu体育馆里面，然后给出m个他们之间的距离， A B X，代表B的座位比A多X. 然后求出这m个关系之间有多少个错误，所谓错误就是当前这个关系与之前的有冲突。

分析与总结：

题目有一句话：we assume the number of rows were infinite. 就是假设这个体育馆是无限大的，所以不用考虑圆圈循环的情况。

用带权并查集做，对于并查集中的每一棵数，树根的距离为0，然后以树根作为参照，每个结点的权值代表与树根的距离。

合并A，B时，假设A，B属于不同的树，那么就要合并这两棵树，把A树合并到B树上，这时要给A树的跟结点root_a赋值，关键是给root_a附上一个什么值。由于A点和B点的权值rank[A]和rank[B]都是相对跟结点的距离，所以分析A，B之间的相对距离，可以得到rank[root_a] = rank[A]+x-rank[B]。注意到这时，对于原来的A的树，只跟新了root_a跟结点的权值，那么其它结点的跟新在查找的那一步里面实行了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>

using namespace std;
const int N=50000+10;
int t,n,m,a,b,x,pre[N],cnt[N];
int find(int x){

    if(x==pre[x]) return pre[x];
    int t=pre[x];
    pre[x] = find(pre[x]);
    cnt[x] += cnt[t];
    return pre[x];
}
bool Union(int x,int y, int m){
    int a=find(x), b=find(y);
    if(a==b){
        if(cnt[x]+m!=cnt[y])
            return false;
        return true;
    }
    pre[b] = a;
    cnt[b] = cnt[x]+m-cnt[y];
    return true;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF){
         for(int i=1;i<=n;i++){
            pre[i]=i;
            cnt[i]=0;
        }
        int r=0;
        while(m--){
            scanf("%d%d%d",&a,&x,&b);
            if(!Union(a,x,b))
                ++r;
        }
        cout << r << endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

题意：这道题的意思是，给你n个人，然后给出m行，每行是两个人的座位，其中座位是从1-n，并且给出两个人的距离，比方说：如果A在4，B在6，那么A和B的距离是2

题解：这道题是一道并查集：

（1）弄清题意，找出出现冲突的位置，判断冲突很简单就是当两个人在同一行坐，同时他们到根节点的距离差值正好是他们之间的差值，此时就出现了冲突了。
（2）关键有两个地方，这也是并查集题目的难点，就是压缩集合，和求节点到根的距离。这里压缩集合就很简单了，一个通用的递归。求到跟的距离dist[a] += dist[tem]; dist[rb]=dist[a]+x-dist[b];注意这两行代码，这是核心代码，首先第一行是求出节点a到根的距离。第二行代码使用的是数学中向量计算的原理如图

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=51000;
int f[maxn],d[maxn];
int n,m;
int find(int x)
{
    if(f[x]==x)
        return x;
    int temp=f[x];
    f[x]=find(f[x]);
    d[x]+=d[temp];
    return f[x];
}
int join(int a,int b,int t1,int t2,int x)
{
    f[t2]=t1;
    d[t2]=d[a]+x-d[b];
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int i,j,a,b,dist,ans=0,cnt=0;
        for(i=0;i<=n;i++)
            f[i]=i,d[i]=0;
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&dist);
            int t1=find(a),t2=find(b);
            if(t1!=t2)
                join(a,b,t1,t2,dist);
            else
            {
                if(d[a]+dist!=d[b])
                ans++;
            }
        }
        printf("%d\n",ans);
    }
}