Bzoj 1703 [Usaco2007 mar]ranking the Cows cow ranking Bitset optimization

Test Instructions:Link Method:Bitset Delivery Closure parsing:Obviously the answer is the number of unordered point pairs. But what about the number of disordered points? The principle of tolerance and repulsion. The answer is the number of all point pairs minus ordered point pairs. How to maintain the number of ordered points? Bitset transitive closures. The first time in life Rnk1!!!!!!!!!!! Code:

#include <bitset>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 1010using namespace STD;intN,m,cnt,cnt2;intHead[n];intV[n]; bitset<N>B[n];structnode{intFrom,to,next;} edge[n*Ten];voidInit () {memset(head,-1,sizeof(head)); Cnt=1;}voidEdgeadd (intFromintTo) {Edge[cnt].from=from;    Edge[cnt].to=to;    Edge[cnt].next=head[from]; head[from]=cnt++;}intRetvoidDfsintx) {v[x]=1; for(inti=head[x];i!=-1; i=edge[i].next) {intto=edge[i].to;if(!v[to]) DFS (to);    B[x]|=b[to]; }}intAnsintMain () {scanf("%d%d", &n,&m); Init (); for(intI=1; i<=n;i++) b[i][i]=1; for(intI=1; i<=m;i++) {intx, y;scanf("%d%d", &x,&y);    Edgeadd (x, y); } for(intI=1; i<=n;i++) {if(!v[i]) DFS (i); } for(intI=1; i<=n;i++) {ans+=b[i].count (); }printf("%d\n", N (n1)/2-ans+n);}

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Bzoj 1703 [Usaco2007 mar]ranking the Cows cow ranking Bitset optimization