LOJ#2305. 「NOI2017」游戏

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 }
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posted @ 2018-06-26 21:44  Blue233333  阅读(175)  评论(0编辑  收藏  举报