鏈接：https://ac.nowcoder.com/acm/contest/1069/K
來源：牛客網

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
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.
輸入描述:

• 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
  輸出描述:
• Line 1: A single integer representing the number of cows whose ranks can be determined
  示例1
  輸入
  復制

5 54 34 23 21 22 5

輸出
復制

2

說明
Cow 2 loses to cows 1, 3, and 4. Thus, cow 2 is no better than any of the cows 1, 3, and 4. Cow 5 loses to cow 2, so cow 2 is better than cow 5. Thus, cow 2 must be fourth, and cow 5 must be fifth. The ranks of the other cows cannot be determ
/*
題意大概是：
給出兩個人的比賽成績優劣，
求最終有多少人的名次（排名）可以確定。
構造圖：
通過構造成績低的指向成績高的單向邊來建圖
我們發現，只要確定每個點有多少個其他的點可達
就可以確定哪些點的名次是可以確定的。
為什么呢？
我們知道根據我們建的圖，
名次越低越少的點可達它，
假設有n個點且所有的點名次都確定，
根據名次由低到高，能到達每個點的個數（除了它本身）一定是唯一確定的：
0，1, 2, 3，4，。。n-1。
所以我們可以根據Floyd求出任意兩點的距離，
根據距離判斷每個點有多少個可達點。
然后我們對每個點可達點個數排一個序，
根據類似上面舉例的序列（0，1, 2, 3，4，。。n-1。）來記錄
可以確定名次的人的個數
*/

#include <bits/stdc++.h> using namespace std; const int N = 105; int a[N][N];//i--j的距離 int cnt[N];//到達i號點的個數（除了本身） int b[N];//輔助排序的數組 int mp[N];//可達點個數的次數，每個能確定名次的點可達點數目的唯一性 int main() {int n,m;cin>>n>>m;for(int i = 1; i <= n; i++){cnt[i] = 0;mp[i] = 0;for(int j = 1; j <= n; j++){if(i==j) a[i][j] = 0;else a[i][j] = -100000;}}int x,y;for(int i = 1; i<= m; i++){cin>>x>>y;a[y][x] = 1;}for(int k = 1; k <= n; k++){for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){a[i][j] = max(a[i][j],a[i][k]+a[k][j]);}}}for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(a[i][j] > 0) cnt[j]++;}}int k = 0;for(int i = 1; i<= n; i++){b[k++] = cnt[i];mp[cnt[i]]++;}sort(b,b+k);int ans = 0;for(int i = 0; i < k; i++){if(b[i]==i&&mp[b[i]]==1)ans++;}cout<<ans<<endl;return 0; } 與50位技術專家面對面20年技術見證，附贈技術全景圖

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