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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> poj3660 Cow Contest的相關內容介紹

poj3660 Cow Contest的相關內容介紹

編輯:C++入門知識

poj3660 Cow Contest的相關內容介紹


Cow Contest   Time Limit:1000MS   Memory Limit:65536K Total Submissions:8986   Accepted:5045  

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 cowAhas a greater skill level than cowB(1 ≤AN; 1 ≤BN;AB), then cowAwill always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM(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:NandM
* 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:AandB

 

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

Sample Output

2

Source

USACO 2008 January Silver  
  題意:n頭牛,給出一部分牛的排名情況,求最後確定排名的牛的頭數。   分析:。。。。坑的一逼,不想說話,我想靜靜。  
 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010

int n,m;
int s[105][105];

int main()
{
    int a,b;
    while(scanf("%d%d",&n,&m)==2)
    {
        CL(s, 0);
        for(int i=0; i


   

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