[bzoj2208][Jsoi2010]连通数

Posted 2020-09-12 FallDream

[bzoj2208][Jsoi2010]连通数

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n<=2000

bitset优化floyd , 枚举k，枚举i，如果i能到k，那么i的bitset直接或上k的。复杂度$O(\\frac{n^{3}}{32})$

#include<iostream>
#include<cstdio>
#include<bitset>
#define MN 2000
using namespace std;
inline int read()
{
    int x = 0 , f = 1; char ch = getchar();
    while(ch < \'0\' || ch > \'9\'){ if(ch == \'-\') f = -1;  ch = getchar();}
    while(ch >= \'0\' && ch <= \'9\'){x = x * 10 + ch - \'0\';ch = getchar();}
    return x * f;
}

int n,ans=0;
char st[MN+5];
bitset<MN+1> b[MN+2];

int main()
{
    n=read();
    for(int i=1;i<=n;i++)
    {
        scanf("%s",st+1);
        for(int j=1;j<=n;j++)
            b[i][j]=(st[j]==\'1\'||i==j)?1:0;    
    }
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            if(b[i][k]) b[i]|=b[k];
    for(int i=1;i<=n;i++) ans+=b[i].count();
    cout<<ans;
    return 0;
}