「CF467D」Fedor and Essay

2014年9月19日3,0970

After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the English teacher told him to prepare an essay. Fedor didn’t want to prepare the essay, so he asked Alex for help. Alex came to help and wrote the essay for Fedor. But Fedor didn’t like the essay at all. Now Fedor is going to change the essay using the synonym dictionary of the English language.

Fedor does not want to change the meaning of the essay. So the only change he would do: change a word from essay to one of its synonyms, basing on a replacement rule from the dictionary. Fedor may perform this operation any number of times.

As a result, Fedor wants to get an essay which contains as little letters «R» (the case doesn’t matter) as possible. If there are multiple essays with minimum number of «R»s he wants to get the one with minimum length (length of essay is the sum of the lengths of all the words in it). Help Fedor get the required essay.

Please note that in this problem the case of letters doesn’t matter. For example, if the synonym dictionary says that word cat can be replaced with word DOG, then it is allowed to replace the word Cat with the word doG.

Input

The first line contains a single integer m (1 ≤ m ≤ 10⁵) — the number of words in the initial essay. The second line contains words of the essay. The words are separated by a single space. It is guaranteed that the total length of the words won’t exceed 10⁵ characters.

The next line contains a single integer n (0 ≤ n ≤ 10⁵) — the number of pairs of words in synonym dictionary. The i-th of the next n lines contains two space-separated non-empty words x_i and y_i. They mean that word x_i can be replaced with word y_i (but not vise versa). It is guaranteed that the total length of all pairs of synonyms doesn’t exceed 5·10⁵ characters.

All the words at input can only consist of uppercase and lowercase letters of the English alphabet.

Output

Print two integers — the minimum number of letters «R» in an optimal essay and the minimum length of an optimal essay.

Sample test(s)

input

3
AbRb r Zz
4
xR abRb
aA xr
zz Z
xr y

AbRb r Zz

xR abRb

aA xr

zz Z

xr y

output

2 6

2 6

input

2
RuruRu fedya
1
ruruRU fedor

RuruRu fedya

ruruRU fedor

output

1 10

1 10

题解

题意是给n个字符串，再给m个转换，要求对n个字符串任意转换，使得最后n个串中r的数量最小，不区分大小写

输出r最小能有几个，满足条件的n个串的最小长度

我的做法是tarjan缩点后在树上dp出每个连通块能转换的最小个数以及长度。。。

我刚锁上就被叉了。。。

当我领悟要开longlong已经迟了。。。

类似这样的数据，给定10w个r

r->10w个a

简直丧病

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<map>
#define inf 1000000000
#define ll long long
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;
}
ll ans1,ans2;
int n,m,cnt,ind,sz,top,scc;
int dfn[300005],low[300005],belong[300005],q[300005];
int id[300005],last[300005],lt[300005];
int len[300005],key[300005],mn[300005],lmn[300005];
bool vis[300005],inq[300005];
char ch[300005];
map<string,int> mp;
struct edge{int to,next;}e[300005],ed[300005];
void insert(int u,int v)
{
	e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;
}
void ins(int u,int v)
{
	ed[++cnt].to=v;ed[cnt].next=lt[u];lt[u]=cnt;
}
void tarjan(int x)
{
	dfn[x]=low[x]=++ind;
	inq[x]=1;q[++top]=x;
	for(int i=last[x];i;i=e[i].next)
		if(!dfn[e[i].to])
			tarjan(e[i].to),low[x]=min(low[x],low[e[i].to]);
		else if(inq[e[i].to])low[x]=min(low[x],dfn[e[i].to]);
	int now=0;
	if(low[x]==dfn[x])
	{
		scc++;
		while(now!=x)
		{
			now=q[top--];
			inq[now]=0;
			belong[now]=scc;
			if(key[now]<mn[scc]||(key[now]==mn[scc]&&len[now]<lmn[scc]))
				mn[scc]=key[now],lmn[scc]=len[now];
		}
	}
}
void rebuild(int t)
{
	cnt=0;
	for(int i=1;i<=t;i++)
		for(int j=last[i];j;j=e[j].next)
			if(belong[i]!=belong[e[j].to])
				ins(belong[i],belong[e[j].to]);
}
void dfs(int x)
{
	if(vis[x])return;vis[x]=1;
	for(int i=lt[x];i;i=ed[i].next)
	{
		int t=ed[i].to;
		dfs(t);
		if(mn[t]<mn[x]||(mn[t]==mn[x]&&lmn[t]<lmn[x]))
			lmn[x]=lmn[t],mn[x]=mn[t];
	}
}
int main()
{
	memset(mn,127/3,sizeof(mn));
	memset(lmn,127/3,sizeof(lmn));
	n=read();
	for(int i=1;i<=n;i++)
	{
		scanf("%s",ch);
		int l=strlen(ch),k=0;
		for(int j=0;j<l;j++)
		{
			if(ch[j]>='A'&&ch[j]<='Z')ch[j]=ch[j]-'A'+'a';
			if(ch[j]=='r')k++;
		}
		int t=mp[ch];if(!t)t=mp[ch]=++sz;len[t]=l;key[t]=k;
		id[i]=t;
	}
	m=read();
	for(int i=1;i<=m;i++)
	{
		scanf("%s",ch);
		int l=strlen(ch),k=0;
		for(int j=0;j<l;j++)
		{
			if(ch[j]>='A'&&ch[j]<='Z')ch[j]=ch[j]-'A'+'a';
			if(ch[j]=='r')k++;
		}
		int x=mp[ch];if(!x)x=mp[ch]=++sz;len[x]=l;key[x]=k;
		scanf("%s",ch);
		l=strlen(ch),k=0;
		for(int j=0;j<l;j++)
		{
			if(ch[j]>='A'&&ch[j]<='Z')ch[j]=ch[j]-'A'+'a';
			if(ch[j]=='r')k++;
		}
		int y=mp[ch];if(!y)y=mp[ch]=++sz;len[y]=l;key[y]=k;
		insert(x,y);
	}
	int t=mp.size();
	for(int i=1;i<=t;i++)if(!dfn[i])tarjan(i);
	rebuild(t);
	for(int i=1;i<=scc;i++)dfs(i);
	for(int i=1;i<=n;i++)
	{
		ans1+=mn[belong[id[i]]],ans2+=lmn[belong[id[i]]];
	}
	printf("%I64d %I64d\n",ans1,ans2);
	return 0;
}

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#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<map>

#define inf 1000000000

#define ll long long

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;

}

ll ans1,ans2;

int n,m,cnt,ind,sz,top,scc;

int dfn[300005],low[300005],belong[300005],q[300005];

int id[300005],last[300005],lt[300005];

int len[300005],key[300005],mn[300005],lmn[300005];

bool vis[300005],inq[300005];

char ch[300005];

map<string,int> mp;

struct edge{int to,next;}e[300005],ed[300005];

void insert(int u,int v)

{

e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;

}

void ins(int u,int v)

{

ed[++cnt].to=v;ed[cnt].next=lt[u];lt[u]=cnt;

}

void tarjan(int x)

{

dfn[x]=low[x]=++ind;

inq[x]=1;q[++top]=x;

for(int i=last[x];i;i=e[i].next)

if(!dfn[e[i].to])

tarjan(e[i].to),low[x]=min(low[x],low[e[i].to]);

else if(inq[e[i].to])low[x]=min(low[x],dfn[e[i].to]);

int now=0;

if(low[x]==dfn[x])

{

scc++;

while(now!=x)

{

now=q[top--];

inq[now]=0;

belong[now]=scc;

if(key[now]<mn[scc]||(key[now]==mn[scc]&&len[now]<lmn[scc]))

mn[scc]=key[now],lmn[scc]=len[now];

}

void rebuild(int t)

{

cnt=0;

for(int i=1;i<=t;i++)

for(int j=last[i];j;j=e[j].next)

if(belong[i]!=belong[e[j].to])

ins(belong[i],belong[e[j].to]);

}

void dfs(int x)

{

if(vis[x])return;vis[x]=1;

for(int i=lt[x];i;i=ed[i].next)

{

int t=ed[i].to;

dfs(t);

if(mn[t]<mn[x]||(mn[t]==mn[x]&&lmn[t]<lmn[x]))

lmn[x]=lmn[t],mn[x]=mn[t];

}

int main()

{

memset(mn,127/3,sizeof(mn));

memset(lmn,127/3,sizeof(lmn));

n=read();

for(int i=1;i<=n;i++)

{

scanf("%s",ch);

int l=strlen(ch),k=0;

for(int j=0;j<l;j++)

{

if(ch[j]>='A'&&ch[j]<='Z')ch[j]=ch[j]-'A'+'a';

if(ch[j]=='r')k++;

}

int t=mp[ch];if(!t)t=mp[ch]=++sz;len[t]=l;key[t]=k;

id[i]=t;

}

m=read();

for(int i=1;i<=m;i++)

{

scanf("%s",ch);

int l=strlen(ch),k=0;

for(int j=0;j<l;j++)

{

if(ch[j]>='A'&&ch[j]<='Z')ch[j]=ch[j]-'A'+'a';

if(ch[j]=='r')k++;

}

int x=mp[ch];if(!x)x=mp[ch]=++sz;len[x]=l;key[x]=k;

scanf("%s",ch);

l=strlen(ch),k=0;

for(int j=0;j<l;j++)

{

if(ch[j]>='A'&&ch[j]<='Z')ch[j]=ch[j]-'A'+'a';

if(ch[j]=='r')k++;

}

int y=mp[ch];if(!y)y=mp[ch]=++sz;len[y]=l;key[y]=k;

insert(x,y);

}

int t=mp.size();

for(int i=1;i<=t;i++)if(!dfn[i])tarjan(i);

rebuild(t);

for(int i=1;i<=scc;i++)dfs(i);

for(int i=1;i<=n;i++)

{

ans1+=mn[belong[id[i]]],ans2+=lmn[belong[id[i]]];

}

printf("%I64d %I64d\n",ans1,ans2);

return 0;

}

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