标签：des c style class blog code
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
标签：des c style class blog code
原文地址：http://www.cnblogs.com/Randolph87/p/3759308.html
