HDU3047 Zjnu Stadium

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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.
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

2

 

 

 

 

正解：带权并查集

解题报告：

　　带权并查集裸题。

　　大概就是比并查集多维护了一个dis数组，表示的含义就是一个到根的距离。

　　每次我路径压缩的时候顺便把父亲节点的距离加到儿子节点上就可以了，有一定像延迟标记。

　　合并的时候，画个图看看就可以发现，如果合并的是x、y，权值为z，集合的代表元素分别为r1，r2，则dis[r2]=dis[x]+z-dis[y]。直接做就可以了。

//It is made by ljh2000
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <complex>
using namespace std;
typedef long long LL;
const int MAXN = 50011;
int n,m,ans,father[MAXN],dis[MAXN];
 
inline int getint(){
    int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
    if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}
 
inline int find(int x){
	if(father[x]==x) return x;
	int t=father[x]; father[x]=find(father[x]);
	dis[x]+=dis[t]; return father[x];
}
 
inline bool check(int x,int y,int z){
	int r1=find(x),r2=find(y);
	if(r1==r2) { if(dis[y]!=dis[x]+z) return false; return true; }
	dis[r2]=dis[x]+z-dis[y];//可以画图看看距离的计算式
	father[r2]=r1;
	return true;
}
 
inline void work(){
	while(scanf("%d%d",&n,&m)!=EOF) {
		for(int i=1;i<=n;i++) father[i]=i,dis[i]=0;
		int x,y,z; ans=0;
		for(int i=1;i<=m;i++) {
			x=getint(); y=getint(); z=getint();
			if(!check(x,y,z)) ans++;
		}
		printf("%d\n",ans);
	}
}
 
int main()
{
    work();
    return 0;
}