Group faces questions from four yuan \ ((i, h_i, j , h_j) \) limit the selection of car models, is not difficult to think of this question use \ (2-SAT \) to solve.
Consider converted to \ (2-SAT \) model, found that in addition to the map \ (x \) , other maps are only two car models can participate, then put both models into two states.
If \ (S_i = A \) , the status is \ (B \) and \ (C \) .
If \ (S_i = B \) , the status is \ (A \) and \ (C \) .
If \ (S_i = C \) , the status is \ (A \) and \ (B \) .
Discussion then four-tuple, set \ (I \) is the input state, \ (^ I \ Prime \) to another state.
If the section \ (I \) field, \ (H_i \) is not available, is not performed even edges.
If the section \ (I \) field, \ (H_i \) is available, in the \ (J \) field, \ (h_j \) is not available, from \ (I \) to \ (i ^ \ prime \) even side, represents not selected \ (I \) .
If both are available, from \ (I \) to \ (J \) connected edge, indicates when the select \ (I \) , then must choose \ (J \) , while the \ (j ^ \ prime \) to \ (i ^ \ prime \) connected edges, if not represented here selected \ (J \) , is not necessarily selected from the group \ (I \) .
Continue to consider how to deal with the map \ (the X-\) , found that the number \ (d \ leqslant8 \) , the data size is small, then we can use \ (dfs \) to enumerate all possible cases again, and then check whether legitimate.
We need only consider the map \ (x \) is equivalent to map \ (a \) and maps \ (b \) in both cases, because at this time has included \ (A, B, C \ ) three models of.
Time complexity is \ (O (2 ^ D (n-m +)) \) .
Still find many of the implementation details, clear to see the code.
\(code:\)
#include<bits/stdc++.h>
#define maxn 1000010
using namespace std;
template<typename T> inline void read(T &x)
{
x=0;char c=getchar();bool flag=false;
while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}
while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
if(flag)x=-x;
}
int n,d,m,x_cnt;
bool flag;
char s[maxn],a[5],b[5],c1[maxn],c2[maxn];
int pos[maxn];
struct node
{
int x,y;
char a,b;
}t[maxn];
struct edge
{
int to,nxt;
}e[maxn];
int head[maxn],edge_cnt;
void add(int from,int to)
{
e[++edge_cnt]=(edge){to,head[from]};
head[from]=edge_cnt;
}
int dfn_cnt,co_cnt,top;
int dfn[maxn],low[maxn],co[maxn],st[maxn];
bool vis[maxn];
void tarjan(int x)
{
dfn[x]=low[x]=++dfn_cnt;
st[++top]=x;
vis[x]=true;
for(int i=head[x];i;i=e[i].nxt)
{
int y=e[i].to;
if(!dfn[y])
{
tarjan(y);;
low[x]=min(low[x],low[y]);
}
else if(vis[y])
low[x]=min(low[x],dfn[y]);
}
if(low[x]==dfn[x])
{
co_cnt++;
int now;
do
{
now=st[top--];
vis[now]=false;
co[now]=co_cnt;
}while(now!=x);
}
}
bool check()
{
for(int i=1;i<=2*n;++i)
if(!dfn[i])
tarjan(i);
for(int i=1;i<=n;++i)
if(co[i]==co[i+n])
return false;
return true;
}
void clear()
{
edge_cnt=dfn_cnt=co_cnt=top=0;
memset(st,0,sizeof(st));
memset(co,0,sizeof(co));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(head,0,sizeof(head));
}
void work()
{
clear();
for(int i=1;i<=m;++i)
{
int x=t[i].x,y=t[i].y;
char a=t[i].a,b=t[i].b;
if(s[x]==a) continue;
if(s[y]==b)
{
add(x+(a==c2[x])*n,x+(a==c1[x])*n);
continue;
}
add(x+(a==c2[x])*n,y+(b==c2[y])*n);
add(y+(b==c1[y])*n,x+(a==c1[x])*n);
}
if(check())
{
flag=true;
for(int i=1;i<=n;i++)
{
if(co[i]<co[i+n]) printf("%c",c1[i]);
else printf("%c",c2[i]);
}
}
}
void dfs(int x)
{
if(flag) return;
if(x==d+1)
{
work();
return;
}
int now=pos[x];
s[now]='A',c1[now]='B',c2[now]='C',dfs(x+1);
s[now]='B',c1[now]='A',c2[now]='C',dfs(x+1);
}
int main()
{
read(n),read(d);
scanf("%s",s+1);
for(int i=1;i<=n;++i)
{
s[i]+='A'-'a';
if(s[i]=='A') c1[i]='B',c2[i]='C';
if(s[i]=='B') c1[i]='A',c2[i]='C';
if(s[i]=='C') c1[i]='A',c2[i]='B';
if(s[i]=='X') pos[++x_cnt]=i;
}
read(m);
for(int i=1;i<=m;++i)
{
read(t[i].x),scanf("%s",a),read(t[i].y),scanf("%s",b);
t[i].a=a[0],t[i].b=b[0];
}
dfs(1);
if(!flag) printf("-1");
return 0;
}