[bzoj2208][Jsoi2010]连通数

来自FallDream的博客，未经允许，请勿转载，谢谢。

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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;
}

 

转载于:https://www.cnblogs.com/FallDream/p/bzoj2208.html