Description
Input
输入数据第一行是图顶点的数量，一个正整数N。 接下来N行，每行N个字符。第i行第j列的1表示顶点i到j有边，0则表示无边。
Output
输出一行一个整数，表示该图的连通数。
Sample Input
3
010
001
100
Sample Output
9
HINT
对于100%的数据，N不超过2000。
看到这题然后马上打了一个tarjan
然后对每一个强连通分量dfs，A了之后感觉有点奇怪，这个复杂度是多少来着，我好像算不出来，果断百度题解
然后大囧。。。。。。怎么好像正解是tarjan+拓扑排序+状态压缩，只搜到了一个和我一样的做法
然后我想到这样做其实可以随随便便卡掉，还是n三方，于是又打了一遍正解，加个拓扑和状态压缩
1 const
2 maxn=200;
3 var
4 first,c,sum,dfn,low,z:array[0..maxn*2]of longint;
5 next,last:array[0..maxn*maxn*2]of longint;
6 flag:array[0..maxn*2]of boolean;
7 f:array[0..maxn,0..maxn]of boolean;
8 n,cnt,tot,ans,time,s:longint;
9
10 procedure insert(x,y:longint);
11 begin
12 inc(tot);
13 last[tot]:=y;
14 next[tot]:=first[x];
15 first[x]:=tot;
16 end;
17
18 procedure dfs(x:longint);
19 var
20 i:longint;
21 begin
22 inc(time);
23 dfn[x]:=time;
24 low[x]:=time;
25 inc(s);
26 z[s]:=x;
27 flag[x]:=true;
28 i:=first[x];
29 while i<>0 do
30 begin
31 if dfn[last[i]]=0 then
32 begin
33 dfs(last[i]);
34 if low[last[i]]<low[x] then low[x]:=low[last[i]];
35 end
36 else
37 if flag[last[i]] and (low[last[i]]<low[x]) then low[x]:=low[last[i]];
38 i:=next[i];
39 end;
40 if low[x]=dfn[x] then
41 begin
42 inc(cnt);
43 while z[s+1]<>x do
44 begin
45 inc(sum[cnt]);
46 c[z[s]]:=cnt;
47 flag[z[s]]:=false;
48 dec(s);
49 end;
50 end;
51 end;
52
53 procedure init;
54 var
55 i,j:longint;
56 cc:char;
57 begin
58 readln(n);
59 for i:=1 to n do
60 begin
61 for j:=1 to n do
62 begin
63 read(cc);
64 if cc=‘1‘ then insert(i,j);
65 end;
66 readln;
67 end;
68 for i:=1 to n do
69 if dfn[i]=0 then dfs(i);
70 for i:=1 to n do
71 begin
72 j:=first[i];
73 while j<>0 do
74 begin
75 if f[c[i],c[last[j]]]=false then
76 begin
77 insert(n+c[i],n+c[last[j]]);
78 f[c[i],c[last[j]]]:=true;
79 end;
80 j:=next[j];
81 end;
82 end;
83 end;
84
85 function dfs2(x:longint):longint;
86 var
87 i:longint;
88 begin
89 dfs2:=sum[x-n];
90 flag[x]:=true;
91 i:=first[x];
92 while i<>0 do
93 begin
94 if flag[last[i]]=false then inc(dfs2,dfs2(last[i]));
95 i:=next[i];
96 end;
97 end;
98
99 procedure work;
100 var
101 i,j:longint;
102 begin
103 for i:=1 to cnt do
104 begin
105 for j:=1 to cnt do
106 flag[j+n]:=false;
107 inc(ans,sum[i]*dfs2(i+n));
108 end;
109 writeln(ans);
110 end;
111
112 begin
113 init;
114 work;
115 end.
1 const
2 maxn=2020;
3 var
4 first,c,sum,dfn,low,z,d:array[0..maxn*2]of longint;
5 next,last:array[0..maxn*maxn*2]of longint;
6 flag:array[0..maxn*2]of boolean;
7 ff:array[0..maxn,0..maxn]of boolean;
8 n,cnt,tot,ans,time,s:longint;
9
10 procedure insert(x,y:longint);
11 begin
12 inc(tot);
13 last[tot]:=y;
14 next[tot]:=first[x];
15 first[x]:=tot;
16 end;
17
18 procedure dfs(x:longint);
19 var
20 i:longint;
21 begin
22 inc(time);
23 dfn[x]:=time;
24 low[x]:=time;
25 inc(s);
26 z[s]:=x;
27 flag[x]:=true;
28 i:=first[x];
29 while i<>0 do
30 begin
31 if dfn[last[i]]=0 then
32 begin
33 dfs(last[i]);
34 if low[last[i]]<low[x] then low[x]:=low[last[i]];
35 end
36 else
37 if flag[last[i]] and (low[last[i]]<low[x]) then low[x]:=low[last[i]];
38 i:=next[i];
39 end;
40 if low[x]=dfn[x] then
41 begin
42 inc(cnt);
43 while z[s+1]<>x do
44 begin
45 inc(sum[cnt]);
46 c[z[s]]:=cnt;
47 flag[z[s]]:=false;
48 dec(s);
49 end;
50 end;
51 end;
52
53 procedure init;
54 var
55 i,j:longint;
56 cc:char;
57 begin
58 readln(n);
59 for i:=1 to n do
60 begin
61 for j:=1 to n do
62 begin
63 read(cc);
64 if cc=‘1‘ then insert(i,j);
65 end;
66 readln;
67 end;
68 for i:=1 to n do
69 if dfn[i]=0 then dfs(i);
70 for i:=1 to n do
71 begin
72 j:=first[i];
73 while j<>0 do
74 begin
75 if (ff[c[i],c[last[j]]]=false) and (c[i]<>c[last[j]]) then
76 begin
77 insert(n+c[i],n+c[last[j]]);
78 inc(d[c[last[j]]]);
79 ff[c[i],c[last[j]]]:=true;
80 end;
81 j:=next[j];
82 end;
83 end;
84 end;
85
86 var
87 q:array[0..maxn]of longint;
88 f:array[0..maxn,0..70]of longint;
89 l,r:longint;
90
91 procedure work;
92 var
93 i,j,k,tmp:longint;
94 begin
95 l:=1;
96 r:=0;
97 for i:=1 to cnt do
98 if d[i]=0 then
99 begin
100 inc(r);
101 q[r]:=i;
102 end;
103 while l<=r do
104 begin
105 j:=first[q[l]+n];
106 while j<>0 do
107 begin
108 dec(d[last[j]-n]);
109 if d[last[j]-n]=0 then
110 begin
111 inc(r);
112 q[r]:=last[j]-n;
113 end;
114 j:=next[j];
115 end;
116 inc(l);
117 end;
118 for i:=r downto 1 do
119 begin
120 f[q[i],q[i] div 30]:=1<<(q[i]mod 30);
121 j:=first[q[i]+n];
122 while j<>0 do
123 begin
124 for k:=0 to cnt div 30 do
125 f[q[i],k]:=f[q[i],k]or f[last[j]-n,k];
126 j:=next[j];
127 end;
128 end;
129 for i:=1 to cnt do
130 begin
131 tmp:=0;
132 for j:=1 to cnt do
133 if f[i,j div 30] and (1<<(j mod 30))>0 then inc(tmp,sum[j]);
134 inc(ans,tmp*sum[i]);
135 end;
136 writeln(ans);
137 end;
138
139 begin
140 init;
141 work;
142 end.
2208: [Jsoi2010]连通数 - BZOJ,布布扣,bubuko.com
时间： 2025-01-03 17:49:28