P2419 [USACO08JAN]牛大赛Cow Contest

P2419 [USACO08JAN]牛大赛Cow Contest

题目背景

[Usaco2008 Jan]

题目描述

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.

FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:)。在赛场上,奶牛们按1..N依次编号。每头奶牛的编程能力不尽相同,并且没有哪两头奶牛的水平不相上下,也就是说,奶牛们的编程能力有明确的排名。 整个比赛被分成了若干轮,每一轮是两头指定编号的奶牛的对决。如果编号为A的奶牛的编程能力强于编号为B的奶牛(1 <= A <= N; 1 <= B <= N; A != B) ,那么她们的对决中,编号为A的奶牛总是能胜出。 FJ想知道奶牛们编程能力的具体排名,于是他找来了奶牛们所有 M(1 <= M <= 4,500)轮比赛的结果,希望你能根据这些信息,推断出尽可能多的奶牛的编程能力排名。比赛结果保证不会自相矛盾。

输入输出格式

输入格式:

第1行: 2个用空格隔开的整数:N 和 M

第2..M+1行: 每行为2个用空格隔开的整数A、B,描述了参加某一轮比赛的奶 牛的编号,以及结果(编号为A,即为每行的第一个数的奶牛为 胜者)

输出格式:

第1行: 输出1个整数,表示排名可以确定的奶牛的数目

输入输出样例

输入样例#1:
5 5
4 3
4 2
3 2
1 2
2 5
输出样例#1:
2

说明

输出说明:

编号为2的奶牛输给了编号为1、3、4的奶牛,也就是说她的水平比这3头奶

牛都差。而编号为5的奶牛又输在了她的手下,也就是说,她的水平比编号为5的

奶牛强一些。于是,编号为2的奶牛的排名必然为第4,编号为5的奶牛的水平必

然最差。其他3头奶牛的排名仍无法确定。

分析:Floyd,建图时有向边,在数一下到这个点的确定了的边,如果除它以外到其他所有的点的距离都确定了,这个点可以确定,ans++;

 #include<cstdio>
#include<cstring>
const int MAXN = ;
int w[MAXN][MAXN];
int cnt[MAXN];
int ans,n,m,flag;
int main()
{
scanf("%d%d",&n,&m);
for (int i=; i<=n; ++i)
for (int j=; j<=n; ++j)
w[i][j] = 1e7;
for (int x,y,i=; i<=m; ++i)
{
scanf("%d%d",&x,&y);
w[y][x] = ;
}
for (int k=; k<=n; ++k)
for (int i=; i<=n; ++i)
for (int j=; j<=n; ++j)
if (w[i][k]+w[k][j]<w[i][j])
w[i][j] = w[i][k]+w[k][j];
for (int i=; i<=n; ++i)
for (int j=; j<=n; ++j)
if (w[i][j]<1e7) cnt[i]++, cnt[j]++;
for (int i=; i<=n; ++i)
if (cnt[i]==n-) ans++;
printf("%d",ans);
return ;
}
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