「2019-8-11提高模拟赛」女装盛宴 (flag)

传送门

Solution 

基环树+倍增+双指针

第一次因为#define int long long而玄学RE

为什么标程都不用开(long long)

Code 

/*玄学RE 看来define ll long long 也有bug*/
#include<bits/stdc++.h>
#define ll long long
#define double
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"
"
#define dbg3(x) cerr<<#x<<"
"
using namespace std;
#define reg register
#define Z(x) memset(x,0,sizeof x)
inline ll read()
{
	ll 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<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
const int MN=2e5+5;
ll n,m,a[MN],b[MN];bool mk[MN];
std::vector<int> G[MN];

ll len,c[MN<<1],la[MN];bool vis[MN];
void getcir(int x)
{
	len=0;int y;
	for(;;la[y]=x,vis[y]=1,x=y)
	{
		y=a[x];
		if(vis[y])
		{
			for(c[++len]=y;x^y;x=la[x])c[++len]=x;
			return;
		}
	}
}

ll d[MN],ans;
void dfs(int x,int fa,int no=0)
{
	int i;mk[x]=1;
	for(i=G[x].size()-1;~i;--i)if((G[x][i]^fa)&&(G[x][i]^no))
	{
		dfs(G[x][i],x);
		if(d[G[x][i]]+b[G[x][i]]>m) ++ans;
		else d[x]=max(d[x],d[G[x][i]]+b[G[x][i]]);
	}
}

ll f[20][MN],q[MN<<1],l,r,s[MN<<1];
void Push(int i){for(;r>=l&&d[c[q[r]]]-s[q[r]]<=d[c[i]]-s[i];--r);q[++r]=i;}
ll cal(ll x,ll y)
{
    ll _=len,tmp=0;
    for(int i=y;~i;--i)
        if(f[i][x]<_)tmp+=1<<i,_-=f[i][x],x=(x+f[i][x]-1)%len+1;
	return tmp+1;
}

void solve(int O)
{
	reg int i,j;
	len=0;getcir(O);
	for(i=1;i<=len;++i)dfs(c[i],0,c[i%len+1]);
	reverse(c+1,c+len+1);
	for(i=1;i<=len;++i)c[i+len]=c[i];
	for(i=2;i<=len*2;++i)s[i]=s[i-1]+b[c[i-1]];

	for(q[l=r=1]=i=j=1;i<=len;++i)
	{
		for(;r>=l&&q[l]<i;++l);if(r<l)q[++r]=++j;
		for(;j<i+len&&s[j]-s[q[l]]+d[c[q[l]]]<=m;++j,Push(j));
		if(j==i+len){++ans;return;}f[0][i]=j-i;
	}
 	ll Add=n,fl=0;
	for(i=1;i<20;++i)
	{
		for(j=1;j<=len;++j)
		{
			f[i][j]=f[i-1][j]+f[i-1][(j+f[i-1][j]-1)%len+1];
			if(f[i][j]>=len-1) Add=min(Add,cal(j,i));
		}
		if(Add^n){ans+=Add;return;}
	}
}

int main()
{
#ifndef LOCAL
	freopen("flag.in","r",stdin);
	freopen("flag.out","w",stdout);
#endif
	int Case=read();
	while(Case--)
	{
		Z(d);ans=0;Z(vis);Z(mk);
		n=read(),m=read();
		reg int i;
		for(i=1;i<=n;++i) G[i].clear();
		for(i=1;i<=n;++i)
			a[i]=read(),b[i]=read(),G[a[i]].push_back(i);
		for(i=1;i<=n;++i)if(!mk[i])solve(i);
		printf("%lld
",ans);
	}
	return 0;
}


Blog来自PaperCloud,未经允许,请勿转载,TKS!

原文地址:https://www.cnblogs.com/PaperCloud/p/11337066.html