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HDOJ3074 Zjnu Stadium(带权并查集)

2018 年 04 月 20 日 • 阅读: 1029 • 数据结构阅读设置

Zjnu Stadium

  • Time Limit: 2000/1000 MS (Java/Others)
  • Memory Limit: 32768/32768 K (Java/Others)
  • Total Submission(s): 4730
  • Accepted Submission(s): 1822

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

PS: the 5th and 10th requests are incorrect

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

Recommend

gaojie


链接

http://acm.hdu.edu.cn/showproblem.php?pid=3047

题意

体育馆的座位布局为圆环形,总共由300列组成,行数不限。现有一个规则:A B X,表明A和B之间必须相差X列。共有n个人和m组判断,问有多少组判断是错误的。

题解

只需要用一个数组sum[x]记录x与其根节点rootx相差的列数,圆环形需要取模300。

x和y进行合并时:

  • F[rooty] = rootx
  • sum[rooty] = (300 + sum[x] - sum[y]) % 300

x进行更新时:

  • sum[x] = (sum[x] + sum[rootx) % 300

判断操作:

  • (300 + sum[y] - sum[x]) % 300 != val

具体参见代码,这一系列的题目其实都是一样的。

代码

StatusAccepted
Time280ms
Memory2072kB
Length988
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 50010;
int F[maxn];
int sum[maxn];

int find(int x)
{
    if (x == F[x])
        return x;
    int tmp = find(F[x]);
    sum[x] = (sum[x] + sum[F[x]]) % 300;
    F[x] = tmp;
    return F[x];
}

int main()
{
    int n, m, x, y, val, ans;
    while (scanf("%d%d", &n, &m) == 2)
    {
        ans = 0;
        for (int i = 0; i <= n; ++i)
        {
            F[i] = i;
            sum[i] = 0;
        }
        for (int i = 0; i < m; ++i)
        {
            scanf("%d%d%d", &x, &y, &val);
            int rootx = find(x), rooty = find(y);
            if (rootx != rooty)
            {
                F[rooty] = rootx;
                sum[rooty] = (300 + sum[x] - sum[y] + val) % 300;
            }
            else
            {
                if ((300 + sum[y] - sum[x]) % 300 != val)
                    ans++;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

The end.
2018-04-20 星期五