长这么大这种题不能1A。。
$n leq 50000$的abcx序列，要你构造同样长一个ABC序列满足：如果这一位是a那就不能填A；如果是b不能填B；如果是c不能填C；否则随意。并满足$m leq 100000$个限制：如果第$i$位填了$p$，那么第$j$位一定要填$q$。$x$个数不超过8.
这不是2-SAT模板？？至于$x$，枚举一下是$a$，是$b$，还是$c$就好了，才怪嘞这样不T死。你想啊，填了$a$不行说明B和C都行不通，那你再找$b$或$c$验证一下A行不行得通不就得了。然后就是2-SAT模板日常写错QAQ
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 //#include<math.h> 5 //#include<set> 6 //#include<queue> 7 //#include<bitset> 8 //#include<vector> 9 #include<algorithm> 10 #include<stdlib.h> 11 using namespace std; 12 13 #define LL long long 14 int qread() 15 { 16 char c; int s=0,f=1; while ((c=getchar())<'0' || c>'9') (c=='-') && (f=-1); 17 do s=s*10+c-'0'; while ((c=getchar())>='0' && c<='9'); return s*f; 18 } 19 20 //Pay attention to '-' , LL and double of qread!!!! 21 22 int n,m; 23 #define maxn 200011 24 struct Edge{int to,next;}edge[maxn<<1],edge2[maxn<<1]; int first[maxn],le=2,first2[maxn],le2; 25 void in(int x,int y) {Edge &e=edge[le]; e.to=y; e.next=first[x]; first[x]=le++;} 26 void in2(int x,int y) {Edge &e=edge2[le2]; e.to=y; e.next=first2[x]; first2[x]=le2++;} 27 28 int tot,Time,top,sta[maxn],bel[maxn],dfn[maxn],low[maxn],opp[maxn],cho[maxn],du[maxn],que[maxn],head,tail; 29 bool insta[maxn]; 30 void tarjan(int x) 31 { 32 dfn[x]=low[x]=++Time; 33 sta[++top]=x; insta[x]=1; 34 for (int i=first[x];i;i=edge[i].next) 35 { 36 Edge &e=edge[i]; 37 if (!dfn[e.to]) tarjan(e.to),low[x]=min(low[x],low[e.to]); 38 else if (insta[e.to]) low[x]=min(low[x],dfn[e.to]); 39 } 40 if (dfn[x]==low[x]) 41 { 42 tot++; 43 while (sta[top]!=x) insta[sta[top]]=0,bel[sta[top]]=tot,top--; 44 insta[sta[top]]=0,bel[sta[top]]=tot,top--; 45 } 46 } 47 void tag(int x) 48 { 49 cho[x]=2; 50 for (int i=first2[x];i;i=edge2[i].next) 51 { 52 Edge &e=edge2[i]; 53 if (!cho[e.to]) tag(e.to); 54 } 55 } 56 57 void clear() 58 { 59 memset(first,0,sizeof(first)); le=2; 60 memset(first2,0,sizeof(first2)); le2=2; 61 tot=Time=top=0; 62 memset(bel,0,sizeof(bel)); 63 memset(dfn,0,sizeof(dfn)); 64 memset(low,0,sizeof(low)); 65 memset(opp,0,sizeof(opp)); 66 memset(cho,0,sizeof(cho)); 67 memset(du,0,sizeof(du)); 68 } 69 70 char s[maxn]; int pos[11],lp=0; 71 struct Eve{int x1,y1,x2,y2;}eve[maxn]; 72 int pp[maxn][2]; 73 74 int getnum(char c) {return c=='A'?0:(c=='B'?1:2);} 75 char getch(int x) {return x==0?'A':(x==1?'B':'C');} 76 77 bool END; 78 bool work() 79 { 80 // cout<<(s+1)<<endl; 81 clear(); 82 for (int i=1;i<=n;i++) if (s[i]=='a') {pp[i][0]=1; pp[i][1]=2;} 83 else if (s[i]=='b') {pp[i][0]=0; pp[i][1]=2;} 84 else if (s[i]=='c') {pp[i][0]=0; pp[i][1]=1;} 85 for (int i=1;i<=m;i++) 86 { 87 int x=eve[i].x1,y=eve[i].x2,xx=eve[i].y1,yy=eve[i].y2; 88 for (int j=0;j<=1;j++) if (pp[x][j]==xx) 89 { 90 bool flag=0; 91 for (int k=0;k<=1;k++) if (pp[y][k]==yy) in(x+j*n,y+k*n),in(y+(k^1)*n,x+(j^1)*n),flag=1; 92 if (!flag) in(x+j*n,x+(j^1)*n); 93 } 94 } 95 // for (int i=1;i<=4;i++) 96 // for (int j=first[i];j;j=edge[j].next) 97 // cout<<i<<' '<<edge[j].to<<endl;cout<<endl; 98 99 for (int i=1;i<=2*n;i++) if (!dfn[i]) tarjan(i); 100 for (int i=1;i<=n;i++) if (bel[i]==bel[i+n]) return 0; 101 102 for (int i=1;i<=n;i++) opp[bel[i]]=bel[i+n],opp[bel[i+n]]=bel[i]; 103 for (int i=1;i<=2*n;i++) 104 for (int j=first[i];j;j=edge[j].next) 105 { 106 Edge &e=edge[j]; 107 if (bel[i]!=bel[e.to]) in2(bel[e.to],bel[i]),du[bel[i]]++; 108 } 109 // for (int i=1;i<=4;i++) cout<<bel[i]<<' ';cout<<endl; 110 head=tail=1; 111 for (int i=1;i<=tot;i++) if (!du[i]) que[tail++]=i; 112 while (head!=tail) 113 { 114 int u=que[head++],v=opp[u]; 115 if (!cho[u]) {cho[u]=1; tag(v);} 116 for (int i=first2[u];i;i=edge2[i].next) 117 { 118 Edge &e=edge2[i]; 119 du[e.to]--; if (!du[e.to]) que[tail++]=e.to; 120 } 121 } 122 for (int i=1;i<=n;i++) if (cho[bel[i]]==1) putchar(getch(pp[i][0])); else putchar(getch(pp[i][1])); 123 return 1; 124 } 125 126 void dfs(int x) 127 { 128 if (END) return; 129 if (x>lp) {if (work()) END=1; return;} 130 s[pos[x]]='a'; dfs(x+1); if (END) return; 131 s[pos[x]]='b'; dfs(x+1); 132 } 133 134 int main() 135 { 136 n=qread(); lp=qread(); lp=0; 137 scanf("%s",s+1); 138 m=qread(); for (int i=1;i<=m;i++) 139 { 140 eve[i].x1=qread(); 141 char c; while ((c=getchar())<'A' || c>'Z'); 142 eve[i].y1=getnum(c); 143 eve[i].x2=qread(); 144 while ((c=getchar())<'A' || c>'Z'); 145 eve[i].y2=getnum(c); 146 } 147 for (int i=1;i<=n;i++) if (s[i]=='x') pos[++lp]=i; 148 dfs(1); 149 if (!END) puts("-1"); 150 return 0; 151 }