"Problem solution": the world line

Question A: World line

Time limit: 1 Sec   Memory Limit: 256 MB

Face questions

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He declined to publicly face the question.

answer

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I had just finished bitset on the exam? ? ? However, when I did not think race is bitset ......

Saying I did not bitset decadent finish ah rush to fill a wave of space and time complexity ......

bitset optimization violence (according to a promise of God that this is the fruit bitset title ??)

Note dfs at the right time if the destination node has been found over the direct use of the current node or the target node bitset bitset value.

There will be $ 6e4 × 6e4 $ bombing space. Opening a $ 6e4 × 3e2 $ is then run to half and half.

As for how to run half, the direct determination and n / 2 can relationship.

Realization look at the code.

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<cmath>
#include<algorithm>
#include<bitset>
#define rint register int
#define read(A) A=init()
#define MAXN 60050
#define MAXM 100050
using namespace std;
inline int init()
{
    int a=0,b=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')b=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){a=(a<<3)+(a<<1)+ch-'0';ch=getchar();}
    return a*b;
}
bitset <MAXN/2> bit[MAXN];
int n,m,mid,du[MAXN];years;longlongread//
 //write
int tot,first[MAXN],v[MAXM],nxt[MAXM];//front-star
bool dev,vis[MAXN];
inline void add(int uu,int vv)
{
    v[++tot]=vv,nxt[tot]=first[uu],first[uu]=tot;
}
inline void dfs(int x)
{
    vis[x]=1;
    if(!dev){if(x<=mid)bit[x].set(x);}
    else    {if(x>mid)bit[x].set(x-mid);}
    for(rint i=first[x];i;i=nxt[i])
    {
        if(!vis[v[i]])dfs(v[i]);
        bit[x]|=bit[v[i]];
    }
    return ;
}
int main()
{
    read(n),read(m);
    for(rint i=1,in_u,in_v;i<=m;++i)
        read(in_u),read(in_v),add(in_u,in_v),du[in_v]++;
    mid=n/2;dev=0;
    for(rint i=1;i<=n;++i)if(!du[i])dfs(i);
    for(rint i=1;i<=n;++i)
    {
        ans+=bit[i].count();
        bit[i].reset();
    }
    memset(vis,0,sizeof(vis));
    dev=1;
    for(rint i=1;i<=n;++i)if(!du[i])dfs(i);
    for(rint i=1;i<=n;++i)ans+=bit[i].count();
    cout<<ans-n-m<<endl;
}

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