2016SDAU课程练习四1023 Problem W

Problem W

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 48   Accepted Submission(s) : 19

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

10 101 2 1503 4 2001 5 2702 6 2006 5 804 7 1508 9 1004 8 501 7 1009 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>

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<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<algorithm>
 #include<iostream>
 #include<queue>
 #include<map>
 #include<math.h>
 using namespace std;
 typedef long long ll;
 //typedef __int64 int64;
 const int maxn = 50015;
 const int inf = 0x7fffffff;
 const double pi=acos(-1.0);
 const double eps = 1e-8;
 
 int dis[ maxn ],fa[ maxn ];
 
 void init( int n ){
     for( int i=0;i<=n;i++ ){
         dis[ i ] = 0;
         fa[ i ] = i;
     }
     return ;
 }
 int find( int x ){
     if( fa[x]==x ) return x;
     int tmp = find( fa[x] );
     dis[ x ] = dis[ x ]+dis[ fa[x] ];
     dis[ x ] = dis[ x ]%300;
     fa[ x ] = tmp;
     return fa[ x ];
 }
 bool unionab( int a,int b,int c ){
     int x = find(a);
     int y = find(b);
     //printf("fa[%d]=%d,fa[%d]=%d,dis[%d]=%d,dis[%d]=%d\n",a,x,b,y,a,dis[a],b,dis[b]);
     if( x==y ){
         if( ((dis[b]-dis[a]+300)%300)!=c ){//b在a的后面，保证是dis[b]-dis[a]
             //printf("test:%d %d\n",a,b);
             return false;
         }
     }
     else{
         dis[ y ] = dis[ a ]+c-dis[ b ]+300;
         dis[y] %= 300;
         fa[ y ] = x;
     }
     return true;
 }
 int main(){
     int n,m;
     while( scanf("%d%d",&n,&m)==2 ){
         int a,b,c;
         init( n );
         int res = 0;
         while( m-- ){
             scanf("%d%d%d",&a,&b,&c);
             bool f = unionab( a,b,c );
             if( f==false ) res++;
         }
         printf("%d\n",res);
     }
     return 0;
 }