hdu6072Logical Chain

Posted 2020-10-03 Macaulish

Kosaraju算法，然後bitset優化

 

主要是學習一下自寫bitset的姿勢

#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define dow(i,l,r) for(int i=r;i>=l;i--)
#define rep0(i,r) for(int i=0;i<r;i++)
#define repedge(i,x) for(int i=cur[x];i>=0;i=e[i].next)
#define maxn 550
#define maxm 100100
#define LL long long
using namespace std;

int tot,p[maxn],n,m,kk;
char s[maxn];

struct Bitset{
    unsigned int v[8];
    void reset()  //clear
    {
        memset(v,0,sizeof(v));
    }
    void set(int x) //set v[x]=1
    {
        v[x>>5]|=1u<<(x&31);
    }
    void flip(int x) // rev
    {
        v[x>>5]^=1u<<(x&31);
    }
    bool ask(int x)  // is == 1?
    {
        return v[x>>5]>>(x&31)&1;
    }
}vis,g[maxn],rg[maxn];
//vis[] =1 not vis,=0 vis
//g[i][j] =1 exist i->j,=0 not exist
//rg[i][j] =1 exist j->i,=0 not exist;

void add(int j,int k)
{
    g[j].flip(k);
    rg[k].flip(j);
}

int nlz(unsigned int x)
{
   int n;
   n = 1;
   if ((x >> 16) == 0) {n = n +16; x = x <<16;}
   if ((x >> 24) == 0) {n = n + 8; x = x << 8;}
   if ((x >> 28) == 0) {n = n + 4; x = x << 4;}
   if ((x >> 30) == 0) {n = n + 2; x = x << 2;}
   n = n - (x >> 31);
   return 31-n;
}

void dfs0(int x)
{
    vis.flip(x);
    rep0(i,8) {
        unsigned int now=vis.v[i]&g[x].v[i];
        while (now) {
            dfs0(i<<5|nlz(now));
            now=vis.v[i]&g[x].v[i];
        }
    }
    p[++tot]=x;
}

void dfs1(int x)
{
    vis.flip(x);
    ++tot;
    rep0(i,8) {
        unsigned int now=vis.v[i]&rg[x].v[i];
        while (now) {
            dfs1(i<<5|nlz(now));
            now=vis.v[i]&rg[x].v[i];
        }
    }
}

int calc()
{
    vis.reset();
    rep0(i,n) vis.set(i);
    tot=0;
    rep0(i,n)
        if (vis.ask(i)) dfs0(i);
    rep0(i,n) vis.set(i);
    int ans=0;
    dow(i,1,n)
        if (vis.ask(p[i])) {
            tot=0;
            dfs1(p[i]);
            ans+=tot*(tot-1)/2;
        }
    return ans;
}

int main()
{
    int tt;
    scanf("%d",&tt);
    while (tt--) {
        scanf("%d %d",&n,&m);
        rep0(i,n) {
            g[i].reset();
            rg[i].reset();
        }
        rep0(i,n) {
            scanf("%s",s);
            rep0(j,n)
                if (s[j]==‘1‘) add(i,j);
        }
        while (m--) {
            scanf("%d",&kk);
            while (kk--) {
                int j,k;
                scanf("%d %d",&j,&k);
                add(j-1,k-1);
            }
            printf("%d\n",calc());
        }
    }
    return 0;
}

View Code

 

另外是一個總結

Bitset
头文件：#include<bitset>
bitset<32> bits;  
b.any() b中是否存在置为1的二进制位？
b.none() b中不存在置为1的二进制位吗？
b.count() b中置为1的二进制位的个数
b.size() b中二进制位数的个数
b[pos] 访问b中在pos处二进制位
b.test(pos) b中在pos处的二进制位置为1么？
b.set() 把b中所有二进制位都置为1
b.set(pos) 把b中在pos处的二进制位置为1
b.reset( ) 把b中所有二进制位都置为0
b.reset( pos ) 把b中在pos处的二进制位置置为0
b.flip( ) 把b中所有二进制位逐位取反
b.flip( pos ) 把b中在pos处的二进制位取反
b.to_ulong( ) 把b中同样的二进制位返回一个unsigned


int __builtin_ffs (unsigned x) 返回x中最后一个1是从右往左第几位
int __builtin_popcount (unsigned x) 返回x中1的个数
int __builtin_ctz (unsigned x) 返回x末尾0的个数(x等于0时未定义)
int __builtin_clz (unsigned x) 返回x中前导0的个数(x等于0时未定义)
int __builtin_parity (unsigned x) 返回x中1的奇偶性

View Code