LOJ#3101. 「JSOI2019」精准预测
对每个点定义(u,t,0/1)为编号为u的人在t时刻是活/死。
当u在t时刻死亡，那他在t+1时刻也必然死亡，(u,1,t)->(u,1,t+1)
当u在t+1时刻活着，那他必然在t时刻也活着，(u,0,t)->(u,0,t-1)
再对两个限制条件分别连边，
限制1：
(t,x,1)→(t+1,y,1)，(t+1,y,0)→(t,x,0)
限制2：
(t,x,0)→(t,y,1),(t,y,0)→(t,x,1)
要求的是对一个点u在T+1时与他一起存活的人。不难转化为一个传递闭包问题。
从以上连边发现，生情况与死情况分别在内部连边，而只有生->死的边没有死->生的边。
所以原图是拓扑图所以我们的问题变成了对于每一个(x,0,T+1)求出它能够到达的所有(y,1,T+1)的状态数。
画出图，不难发现大量出度为一的点无用，于是可以把这些点全部去掉，也就等价于只留下T+1时刻的点与限制边的出点。
此时只有(2n+2m)个点，于是用bitset优化一下即可。。。
代码：
1 //Copyright@3_soon,From Hangzhou No.2 High School Baimahu 2 //Problem: 3 //Result: 4 #pragma GCC optimize(3) 5 #include<bits/stdc++.h> 6 using namespace std; 7 #define re register 8 #define LL long long 9 #define DB double 10 #define il inline 11 #define For(x,a,b) for(re int x=a;x<=b;x++) 12 #define For2(x,a,b) for(re int x=a;x>=b;x--) 13 #define LFor(x,a,b) for(re LL x=a;x<=b;x++) 14 #define LFor2(x,a,b) for(re LL x=a;x>=b;x--) 15 #define Abs(x) ((x>0)? x:-x) 16 #define pii pair<int,int> 17 #define mp(x,y) make_pair(x,y) 18 #define mod (LL)1000000007 19 #define base 10500 20 int gi() 21 { 22 int res=0,fh=1;char ch=getchar(); 23 while((ch>'9'||ch<'0')&&ch!='-') ch=getchar(); 24 if(ch=='-') fh=-1,ch=getchar(); 25 while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar(); 26 return fh*res; 27 } 28 LL gl() 29 { 30 LL res=0,fh=1;char ch=getchar(); 31 while((ch>'9'||ch<'0')&&ch!='-') ch=getchar(); 32 if(ch=='-') fh=-1,ch=getchar(); 33 while(ch>='0'&&ch<='9')res=res*10+ch-'0',ch=getchar(); 34 return fh*res; 35 } 36 namespace IO{ 37 #define BUF_SIZE 100000 38 #define OUT_SIZE 100000 39 #define ll long long 40 //fread->read 41 42 bool IOerror=0; 43 inline char nc(){ 44 static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE; 45 if (p1==pend){ 46 p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin); 47 if (pend==p1){IOerror=1;return -1;} 48 //{printf("IO error!\n");system("pause");for (;;);exit(0);} 49 } 50 return *p1++; 51 } 52 inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';} 53 inline void read(int &x){ 54 bool sign=0; char ch=nc(); x=0; 55 for (;blank(ch);ch=nc()); 56 if (IOerror)return; 57 if (ch=='-')sign=1,ch=nc(); 58 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 59 if (sign)x=-x; 60 } 61 inline void read(ll &x){ 62 bool sign=0; char ch=nc(); x=0; 63 for (;blank(ch);ch=nc()); 64 if (IOerror)return; 65 if (ch=='-')sign=1,ch=nc(); 66 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 67 if (sign)x=-x; 68 } 69 inline void read(double &x){ 70 bool sign=0; char ch=nc(); x=0; 71 for (;blank(ch);ch=nc()); 72 if (IOerror)return; 73 if (ch=='-')sign=1,ch=nc(); 74 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 75 if (ch=='.'){ 76 double tmp=1; ch=nc(); 77 for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0'); 78 } 79 if (sign)x=-x; 80 } 81 inline void read(char *s){ 82 char ch=nc(); 83 for (;blank(ch);ch=nc()); 84 if (IOerror)return; 85 for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch; 86 *s=0; 87 } 88 inline void read(char &c){ 89 for (c=nc();blank(c);c=nc()); 90 if (IOerror){c=-1;return;} 91 } 92 //fwrite->write 93 struct Ostream_fwrite{ 94 char *buf,*p1,*pend; 95 Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;} 96 void out(char ch){ 97 if (p1==pend){ 98 fwrite(buf,1,BUF_SIZE,stdout);p1=buf; 99 } 100 *p1++=ch; 101 } 102 void print(int x){ 103 static char s[15],*s1;s1=s; 104 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 105 while(x)*s1++=x%10+'0',x/=10; 106 while(s1--!=s)out(*s1); 107 } 108 void println(int x){ 109 static char s[15],*s1;s1=s; 110 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 111 while(x)*s1++=x%10+'0',x/=10; 112 while(s1--!=s)out(*s1); out('\n'); 113 } 114 void print(ll x){ 115 static char s[25],*s1;s1=s; 116 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 117 while(x)*s1++=x%10+'0',x/=10; 118 while(s1--!=s)out(*s1); 119 } 120 void println(ll x){ 121 static char s[25],*s1;s1=s; 122 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 123 while(x)*s1++=x%10+'0',x/=10; 124 while(s1--!=s)out(*s1); out('\n'); 125 } 126 void print(double x,int y){ 127 static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000, 128 1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL, 129 100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL}; 130 if (x<-1e-12)out('-'),x=-x;x*=mul[y]; 131 ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1; 132 ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2); 133 if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);} 134 } 135 void println(double x,int y){print(x,y);out('\n');} 136 void print(char *s){while (*s)out(*s++);} 137 void println(char *s){while (*s)out(*s++);out('\n');} 138 void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}} 139 ~Ostream_fwrite(){flush();} 140 }Ostream; 141 inline void print(int x){Ostream.print(x);} 142 inline void println(int x){Ostream.println(x);} 143 inline void print(char x){Ostream.out(x);} 144 inline void println(char x){Ostream.out(x);Ostream.out('\n');} 145 inline void print(ll x){Ostream.print(x);} 146 inline void println(ll x){Ostream.println(x);} 147 inline void print(double x,int y){Ostream.print(x,y);} 148 inline void println(double x,int y){Ostream.println(x,y);} 149 inline void print(char *s){Ostream.print(s);} 150 inline void println(char *s){Ostream.println(s);} 151 inline void println(){Ostream.out('\n');} 152 inline void flush(){Ostream.flush();} 153 #undef ll 154 #undef OUT_SIZE 155 #undef BUF_SIZE 156 }; 157 using namespace IO; 158 bitset<base>f[600005],g; 159 vector<int>p[50005][2],tim[50005]; 160 int xx[100005],yy[100005],opt[100005]; 161 int t[100005]; 162 int n,m,T; 163 int cnt,tot,ans[100005]; 164 int vis[500005]; 165 int flag[100005]; 166 struct pp{int to,nxt;}e[700005]; 167 int head[500005]; 168 void addedge(int x,int y){e[++tot]=(pp){y,head[x]};head[x]=tot;} 169 int getpos(int x,int y) 170 { 171 int l,r; 172 l=0,r=tim[x].size()-1; 173 while(l<=r) 174 { 175 re int mid=l+r>>1; 176 if(tim[x][mid]==y) return mid; 177 if(tim[x][mid]<y) l=mid+1; 178 else r=mid; 179 } 180 return 0; 181 } 182 void sfd(int u) 183 { 184 if(vis[u]) return; 185 vis[u]=1; 186 for(int i=head[u];i;i=e[i].nxt) 187 sfd(e[i].to),f[u]|=f[e[i].to]; 188 } 189 void solve(int l,int r) 190 { 191 g.reset(); 192 if(l>n) return ; 193 r=min(r,n); 194 For(i,1,cnt) vis[i]=0,f[i].reset(); 195 For(i,l,r) For(j,0,tim[i].size()-1) f[p[i][1][j]][i-l]=1; 196 For(i,1,n) sfd(p[i][0][tim[i].size()-1]); 197 For(i,l,r) 198 { 199 if(f[p[i][0][tim[i].size()-1]][i-l]) g[i-l]=1,flag[i]=1; 200 } 201 For(i,1,n) 202 { 203 if(flag[i]) continue; 204 int u=p[i][0][tim[i].size()-1]; 205 ans[i]+=(f[u]|g).count(); 206 } 207 } 208 209 210 int main() 211 { 212 tot=cnt=0; 213 memset(ans,0,sizeof(ans)); 214 memset(head,-1,sizeof(head)); 215 memset(flag,0,sizeof(flag)); 216 read(T),read(n),read(m); 217 For(i,1,m) 218 { 219 read(opt[i]),read(t[i]),read(xx[i]),read(yy[i]); 220 tim[xx[i]].push_back(t[i]); 221 tim[yy[i]].push_back(t[i]+(opt[i]^1)); 222 } 223 For(i,1,n) 224 { 225 tim[i].push_back(T+1); 226 sort(tim[i].begin(),tim[i].end()); 227 tim[i].erase(unique(tim[i].begin(),tim[i].end()),tim[i].end()); 228 For(j,0,tim[i].size()-1) 229 p[i][0].push_back(++cnt),p[i][1].push_back(++cnt); 230 } 231 For(i,1,n) 232 { 233 For(j,1,tim[i].size()-1) 234 { 235 addedge(p[i][0][j],p[i][0][j-1]); 236 addedge(p[i][1][j-1],p[i][1][j]); 237 } 238 } 239 For(i,1,m) 240 { 241 int u=getpos(xx[i],t[i]),v; 242 if(!opt[i]) 243 { 244 v=getpos(yy[i],t[i]+1); 245 addedge(p[xx[i]][1][u],p[yy[i]][1][v]);addedge(p[yy[i]][0][v],p[xx[i]][0][u]); 246 } 247 else 248 { 249 v=getpos(yy[i],t[i]); 250 addedge(p[xx[i]][0][u],p[yy[i]][1][v]);addedge(p[yy[i]][0][v],p[xx[i]][1][u]); 251 } 252 } 253 For(i,1,5) solve(base*(i-1)+1,base*i); 254 For(i,1,n) printf("%d ",flag[i]? 0:n-(ans[i]+1)); 255 puts(""); 256 return 0; 257 }
