博客
关于我
poj 3660 (floyd)
阅读量:353 次
发布时间:2019-03-04

本文共 2316 字,大约阅读时间需要 7 分钟。

题目

Cow Contest
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 19773 Accepted: 10979
Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1…N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

  • Line 1: Two space-separated integers: N and M
  • Lines 2…M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

  • Line 1: A single integer representing the number of cows whose ranks can be determined

Sample Input

5 5

4 3
4 2
3 2
1 2
2 5
Sample Output

2

分析

碎碎念qwq

最开始我想到的是用拓扑排序,不过迟迟没有思路,看了后豁然开朗awa弗洛伊德的千种套路

正文

题目让我们寻找哪些牛的排名是可以确定的

通过题中给出的案例分析,我们可以得到这样一个关系图

可以看到,2和5的排名是确定的,而1,4,3这三头牛,因为1和4,3的关系不明确,所以这三头牛的关系不确定。

我们又可以从这个图中知道,确定了排名的牛,与其他牛的关系都是确定的,
1,4,3都能打败2,2能打败5
1,2,3,4都能打败5
但是对于1,你只能知道1能打败2,5,但是你不知道3,4和1的关系
所以,我们可以像存图一样,把这个关系存起来
这是一个有向无环图
之后我们就可以用弗洛伊德来更新所有节点之间的关系,找出与其他节点关系确定的节点,就得出最后结果了

代码

#include 
#include
#include
#define mt(x, k) memset(x,k, sizeof(x))using namespace std;int main() { ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); int mp[110][110], n, m, a, b; mt(mp, 0); cin >> n >> m; for (int i = 0; i < m; ++i) { cin >> a >> b; mp[a][b] = 1; } for (int k = 1; k <= n; ++k) { for (int i = 1; i <= n; ++i) { for (int j = 1; j <= n; ++j) { if (mp[i][k] && mp[k][j])mp[i][j] = 1; } } } int ans = 0, j; for (int i = 1; i <= n; ++i) { for (j = 1; j <= n; ++j) { if (i == j)continue; if (mp[i][j] == 0 && mp[j][i] == 0)break; } if (j > n)ans++; } cout << ans << '\n'; return 0;}

转载地址:http://nxir.baihongyu.com/

你可能感兴趣的文章
MySQL Error Handling in Stored Procedures---转载
查看>>
MVC 区域功能
查看>>
MySQL FEDERATED 提示
查看>>
mysql generic安装_MySQL 5.6 Generic Binary安装与配置_MySQL
查看>>
Mysql group by
查看>>
MySQL I 有福啦,窗口函数大大提高了取数的效率!
查看>>
mysql id自动增长 初始值 Mysql重置auto_increment初始值
查看>>
MySQL in 太多过慢的 3 种解决方案
查看>>
MySQL InnoDB 三大文件日志,看完秒懂
查看>>
Mysql InnoDB 数据更新导致锁表
查看>>
Mysql Innodb 锁机制
查看>>
MySQL InnoDB中意向锁的作用及原理探
查看>>
MySQL InnoDB事务隔离级别与锁机制深入解析
查看>>
Mysql InnoDB存储引擎 —— 数据页
查看>>
Mysql InnoDB存储引擎中的checkpoint技术
查看>>
Mysql InnoDB存储引擎中缓冲池Buffer Pool、Redo Log、Bin Log、Undo Log、Channge Buffer
查看>>
MySQL InnoDB引擎的锁机制详解
查看>>
Mysql INNODB引擎行锁的3种算法 Record Lock Next-Key Lock Grap Lock
查看>>
mysql InnoDB数据存储引擎 的B+树索引原理
查看>>
mysql innodb通过使用mvcc来实现可重复读
查看>>