P2881 [Usaco07mar] ranked cattle ranking the cows

Bitset Optimizing transitive closure template problem

This relationship is modeled directly on graph theory, which is actually a transitive closure.

Transitive closures There is a simple way to do this is to Floyd.

And the scope of the problem is \ (n \leq 1000\),\ (n^3\) The violence will obviously t.

And the use of bitset, heard can be optimized to the original method of \ (\frac{1}{32}\) even better!

Directly to the code is the fact that you don't understand the principle

#include<cstdio>#include<bitset>const int maxn = 1005;std::bitset<maxn> b[maxn];int n, m;int main(){    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; i++) b[i][i] = true;    while(m--)    {        int u, v; scanf("%d%d", &u, &v);        b[u][v] = true;    }    for(int k = 1; k <= n; k++)    {        for(int i = 1; i <= n; i++)        {            if(b[i][k])            {                b[i] |= b[k];            }        }    }    int ans = 0;    for(int i = 1; i <= n; i++) ans += b[i].count();    ans -= n;    ans = n * (n - 1) / 2 - ans;    printf("%d\n", ans);    return 0;}

P2881 [Usaco07mar] ranked cattle ranking the cows