ICPCCamp 2016 Day1 ftiasch's Contest #4(All Pair Shortest Path-位运算)

题意：给一张2000个点的有向图（边的权制均为1），你需要求∑ni=1∑nj=1dis(i,j)\sum_{i=1}^n \sum_{j=1}^n dis(i,j) dis(i,j)是i到j的最短路距离。

不要被复杂度骗了，

位运算暴力过，

O(n3/64)O(n^3/64)

#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <functional>
#include <cstdlib>
#include <queue>
#include <stack>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (500000000000LL)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (2000+10)
int n;
ll ans=0;
ll e[MAXN][MAXN];
ll cur[MAXN]={0},old[MAXN],old2[MAXN];
int To[MAXN];
void bfs(int x){
int nc=1,deep=0,sum=1;
Rep(i,40) cur[i]=old[i]=0;
cur[x/60]= old[x/60] = 1LL << (x%60);
while (nc)
{
++deep;
int k=0;
Rep(i,40) Rep(j,60) if (cur[i]>>j & 1) {
To[k++] = i*60 + j;
}
Rep(i,40) old2[i] = old[i] ;
Rep(i,k) Rep(j,40) {
old[j] |= e[To[i]][j];
}
nc=0;
Rep(i,40) {
cur[i]= old[i] ^ old2[i];
nc += __builtin_popcountll((ull)cur[i]);
}
ans+= 1LL * deep * deep * nc;
sum+=nc;
//        cout<<deep << ' '<< nc<<endl;
//        Rep(i,40) cout<<cur[i];

}
ans +=1LL *n*n*(n - sum) ;
}
char s[MAXN];
int main() {
cin >> n;

Rep(i,n) {
cin>>s;
Rep(j,40) {
int t=j*60;
for(int k=0;k<60;k++) {
if (t+k<n && s[t+k] == '1' ) e[i][j] |= 1LL << k;
}
}
}
//    cout<<"1"<<endl;
Rep(i,n) bfs(i);
cout<<ans<<endl;
return 0;
}