题意：

观众席围成一圈。列的总数是300，编号为1–300，顺时针计数，我们假设行的数量是无限的。将有N个人去那里。他对这些座位提出了要求：这意味着编号A的顺时针X距离坐着编号B。例如：A在第4列，X是2，那么B必须在第6列（6=4+2）。现在你的任务是判断请求是否正确。

题目：

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

Hint

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

思路

1.由于题中明确给出，编号A的顺时针X距离坐着编号B，故x<300，当我们用并查集找到一个祖宗节点时，我们可以将该点看作原点，将圈拉成直线看待。即可忽略成环。

2.b->a, a离”原点更近“

int f(int x)
{if(x!=dp[x]){int t=dp[x];/*因为在递归找祖先的过程中，dp[x]的值会改变，而我们我们只需要找到距离x最近的点，即可得到dp[x]到祖宗节点的距离。所以要记录t的值*/dp[x]=f(dp[x]);num[x]+=num[t];}return dp[x];
}

dp[b]->dp[a]

  dp[v]=u;num[v]=num[a]+x-num[b];

AC代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int M=5e4+10;
int dp[M],num[M];
int f(int x)
{if(x!=dp[x]){int t=dp[x];dp[x]=f(dp[x]);/*因为在递归找祖先的过程中，dp[x]的值会改变，而我们我们只需要找到距离x最近的点，即可得到dp[x]到祖宗节点的距离。所以要记录t的值*/num[x]+=num[t];}return dp[x];
}
bool dfs(int a,int b,int x)
{int u=f(a),v=f(b);/*a->u,b->v*/if(u==v){if(num[a]+x!=num[b])/*b->a,又起点为父节点*/return true;return false;}dp[v]=u;num[v]=num[a]+x-num[b];return false;
}
int main()
{int m,n;while(~scanf("%d%d",&m,&n)){int ans=0;for(int i=0;i<m;i++){dp[i]=i;num[i]=0;}while(n--){int a,b,x;/*设起点为祖宗节点，则都指向起点b->a*/scanf("%d%d%d",&a,&b,&x);if(dfs(a,b,x))ans++;}printf("%d\n",ans);}return 0;
}

Zjnu Stadium HDU - 304 加权并查集相关推荐

hdu 3635 Dragon Balls（加权并查集）2010 ACM-ICPC Multi-University Training Contest（19）
这道题说,在很久很久以前,有一个故事.故事的名字叫龙珠.后来,龙珠不知道出了什么问题,从7个变成了n个. 在悟空所在的国家里有n个城市,每个城市有1个龙珠,第i个城市有第i个龙珠. 然后,每经过一段时 ...
Rochambeau POJ - 2912 （枚举和加权并查集+路径压缩）找唯一裁判
题意:有n个人玩石头剪刀布,有且只有一个裁判.除了裁判每个人的出拳形式都是一样的. a<b表示b打败a,a=b表示a和b出拳一样,平手.a>b表示a打败b. 给出m个回合的游戏结果,问能否 ...
A Bug‘s Life POJ 2492 加权并查集
A Bug's Life POJ 2492 加权并查集传送门:http://poj.org/problem?id=2492 Description Background Professor Hopp ...
HDU 3234 Exclusive-OR [并查集]
http://acm.hdu.edu.cn/showproblem.php?pid=3234 #Description 给你N个数,X0-X(N-1) 执行Q个查询三种格式 I p v Xp= v ...
hdu 1232 经典并查集应用
http://acm.hdu.edu.cn/showproblem.php?pid=1232 完全就是并查集的应用啊... View Code 1 #include<iostream> 2 ...
畅通工程 hdu 1232 HDU - 1863 （并查集+最小生成树)
畅通工程hdu 1232 并查集 Problem Description Input Output 参考代码 HDU - 1863 Problem Description Input Output 参 ...
HDU 3234 Exclusive-OR(并查集)
转载请注明出处,谢谢http://blog.csdn.net/acm_cxlove/article/details/7854526 by---cxlove 题目:给出N个数,给出一些条件, ...
HDU 5606 tree 并查集
tree 把每条边权是1的边断开,发现每个点离他最近的点个数就是他所在的连通块大小. 开一个并查集,每次读到边权是0的边就合并.最后Ansi=size[findset(i)],size表示每个并 ...
hdu 1116 欧拉回路并查集一组字符串能否首尾相连成一个字符串
主要是欧拉回路的基础知识,用并查集加工处理注意欧拉回路和并查集的细节判断不能粘贴复制,一定要理解之后再敲一遍代码,否则浪费更多的时间 #include <stdio.h> #inclu ...

Zjnu Stadium HDU - 304 加权并查集

题意：

题目：