省选校内模拟D6

T1

 Solution

• bitset优化套路
• bitset赋值类似数组
• bitset相关

• 关于分块
• 用时间换空间
• 把n个点分成若干个块
• 每次只考虑当前块内的点到达所有点的情况

Code

---

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
using namespace std;
const int N=1e5+10;
struct node{int y,n;}e[N<<1],E[N<<1],ans[N<<1];
int lin[N<<1],LIN[N<<1],v[N],dfn[N],deg[N],inv[N],low[N],c[N],d[N],a[N],b[N],tot=0,cnt=0,LEN=0,len=0,t,n,m,q,x,y;
stack<int> S;
bitset<10000> f[100010];
void tarjan(int x){
	dfn[x]=low[x]=++tot;S.push(x);inv[x]=1;
	for(int i=lin[x];i;i=e[i].n){
		int y=e[i].y;
		if(!dfn[y]){
			tarjan(y);
			low[x]=min(low[x],low[y]);
		}else if(inv[y]) low[x]=min(low[x],low[y]);
	}if(low[x]==dfn[x]){
		cnt++;int y;
		while(x!=y){
			y=S.top();S.pop();inv[y]=0;
			c[y]=cnt;
		}
	}
}
void read1(int x,int y)
{e[++len].y=y,e[len].n=lin[x],lin[x]=len;}
void read2(int x,int y)
{E[++LEN].y=y,E[LEN].n-LIN[x],LIN[x]=LEN;}
bool cmp(node x,node y){
	return x.y<y.y||x.y==y.y&&x.n<y.n;
}
void topsort(int l,int r){
	queue<int> q;
	rep(i,1,cnt){deg[i]=d[i];f[i].reset();}
	rep(i,1,cnt)if(!deg[i])q.push(i);
	while(q.size()){
		int x=q.front();q.pop();
		if(x>=l&&x<=r)f[x][x-l]=1;
		for(int i=LIN[x];i;i=E[i].n){
			int y=E[i].y;
			deg[y]--;
			f[y]|=f[x];
			if(deg[y]==0){
				q.push(y);
			}
		}
	}
	rep(i,1,t){
		if(b[i]>=l&&b[i]<=r&&f[a[i]][b[i]-l]==1){
			puts("NO");exit(0);
		}
	}
}
int main()
{
	freopen("gplt.in","r",stdin);
	freopen("gplt.out","w",stdout);
	scanf("%d%d",&n,&m);
	rep(i,1,m){
		scanf("%d%d",&x,&y);
		read1(x,y);
		ans[i].y=x,ans[i].n=y;
	}
	rep(i,1,n)if(!dfn[i])tarjan(i);
	rep(x,1,n){
		for(int i=lin[x];i;i=e[i].n){
			int y=e[i].y;
			if(c[x]==c[y])continue;
			read2(c[y],c[x]);
			d[c[x]]++;
		}
	}
	scanf("%d",&t);
	rep(i,1,t){
		scanf("%d%d",&x,&y);
		a[i]=c[x];
		b[i]=c[y];
	}
	for(int l=1;l<=cnt;l+=8000){
		int r=min(l+8000,cnt);
		topsort(l,r);
	}
	puts("YES"); printf("%d\n",m);
	rep(i,1,m)printf("%d %d\n",ans[i].y,ans[i].n);
	return 0;
}