标签：ssi push lan efi line cst arm rod ops
Description
Each of Farmer John‘s N cows (1 ≤ N ≤ 1,000) produces milk at a different positive rate, and FJ would like to order his cows according to these rates from the fastest milk producer to the slowest.
FJ has already compared the milk output rate for M (1 ≤ M ≤ 10,000) pairs of cows. He wants to make a list of C additional pairs of cows such that, if he now compares those C pairs, he will definitely be able to deduce the correct ordering of all N cows. Please help him determine the minimum value of C for which such a list is possible.
Input
Output
Sample Input
5 5 2 1 1 5 2 3 1 4 3 4
Sample Output
3
Hint
1 #include<cstdio> 2 #include<cstdlib> 3 #include<stack> 4 #include<bitset> 5 #include<vector> 6 using namespace std; 7 const int N=1024; 8 stack<int> S; 9 bitset<N> a[N]; 10 vector<int> G[N]; 11 int n,m; 12 int d[N]; 13 int read(){ 14 char c;int x=0,f=1; 15 for(c=getchar();!isdigit(c);c=getchar())if(c==‘-‘)f=-1; 16 for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48); 17 return x*f; 18 } 19 void TopSort(){ 20 for(int i=0;i<n;++i) 21 if(!d[i])S.push(i); 22 while(!S.empty()){ 23 int u=S.top();S.pop(); 24 for(int i=0,i_end=G[u].size();i<i_end;++i){ 25 int v=G[u][i]; 26 a[v]|=a[u]; 27 --d[v]; 28 if(!d[v])S.push(v); 29 } 30 } 31 } 32 int main(){ 33 scanf("%d%d",&n,&m); 34 while(m--){ 35 int u=read()-1,v=read()-1; 36 G[u].push_back(v); 37 ++d[v]; 38 } 39 for(int i=0;i<n;++i)a[i][i]=1; 40 TopSort(); 41 for(int i=0;i<n;++i) 42 m+=a[i].count(); 43 printf("%d\n",n*(n-1)/2-m-1+n); 44 return 0; 45 }
就当是对ZJOI2017Day2的交代吧
POJ3275 [USACO07MAR]Ranking the Cows
标签：ssi push lan efi line cst arm rod ops
原文地址：http://www.cnblogs.com/ndqzhang1111/p/6915938.html
