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*HDU3047 并查集

2024-08-18 13:59:12   223人阅读

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3062    Accepted Submission(s): 1182


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

 

Source
2009 Multi-University Training Contest 14 - Host by ZJNU
 
题意：
在一个周长为300的环内，共n个点，每次给出两点之间的顺时针距离，问在前面给出的数据条件下新给出的数据是否正确，统计错误个数。
代码：

技术分享

 1 //num[b]=num[x]+z-num[y] 这个看看图就好理解了
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 using namespace std;
 7 int n,m;
 8 int fat[50004];
 9 int num[50004];
10 int find(int x)
11 {
12     if(fat[x]!=x)
13     {
14         int f=fat[x];
15         fat[x]=find(fat[x]);
16         num[x]+=num[f];
17     }
18     return fat[x];
19 }
20 void connect(int x,int y,int z)
21 {
22     int a=find(x),b=find(y);
23     if(a!=b)
24     {
25         fat[b]=a;
26         num[b]=num[x]+z-num[y];
27     }
28 }
29 int main()
30 {
31     int a,b,c,ans;
32     while(scanf("%d%d",&n,&m)!=EOF)
33     {
34         ans=0;
35         for(int i=0;i<=n;i++)
36         {
37             fat[i]=i;
38             num[i]=0;
39         }
40         for(int i=1;i<=m;i++)
41         {
42             scanf("%d%d%d",&a,&b,&c);
43             if(find(a)==find(b))
44             {
45                 if(num[b]-num[a]!=c)
46                 ans++;
47             }
48             else connect(a,b,c);
49         }
50         printf("%d\n",ans);
51     }
52     return 0;
53 }

 

*HDU3047 并查集


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