POJ 3694

POJ 3694

题目大意:给出一副联通的无向图,求加上一条边后还有多少条桥,多次加边。

解:其实就是tarjan求桥加lca,这里可以朴素记录深度,向上回溯,取消桥的标记即可。我挺二的,首先把最少边使得不存在双连通分量和加边取消双连通分量搞反了,然后去看了lca和rmq,lca不是很懂tarjan的做法,有空去看下。然后就是做题的时候记录了哪些边是桥,wa,其实只要记录点即可,如果一个点是桥标记,则说明他到他父亲这条边是桥,ac。然后_gxx的做法很神奇,首先遍历出一棵树,然后把剩下的原始边加上去,lca,覆盖桥标记(一开始的全是桥),然后新加的边也是这样处理即可。还有就是桥标记的判断是dfn[x]<low[dest]而非low[x]<low[dest],这个是个wa点。

View Code
  1 //poj 3694
2 const
3 maxn=111111;
4 maxm=211111;
5 type
6 data=record
7 dest, next: longint;
8 end;
9 var
10 edge: array[1..maxm*2]of data;
11 pre, vect, dfn, low, dep: array[1..maxn]of longint;
12 brid, visit: array[1..maxn]of boolean;
13 test, n, m, tot, ans, time, q: longint;
14 procedure add(x, y: longint);
15 begin
16 inc(tot);
17 with edge[tot] do begin
18 dest := y;
19 next := vect[x];
20 vect[x] := tot;
21 end;
22 inc(tot);
23 with edge[tot] do begin
24 dest := x;
25 next := vect[y];
26 vect[y] := tot;
27 end;
28 end;
29
30 procedure init;
31 var
32 i, x, y: longint;
33 begin
34 tot := 0; ans := 0;
35 fillchar(vect, sizeof(vect), 0);
36 for i := 1 to m do begin
37 readln(x, y);
38 add(x, y);
39 end;
40 end;
41
42 procedure tarjan(x: longint);
43 var
44 i: longint;
45 flag: boolean;
46 begin
47 visit[x] := true;
48 inc(time);
49 dfn[x] := time; low[x] := time;
50 i := vect[x]; flag := true;
51 while i<>0 do
52 with edge[i] do begin
53 if (pre[x]=dest)and(flag) then begin
54 i := next; flag := false;
55 continue;
56 end;
57 if dfn[dest]=0 then begin
58 pre[dest] := x;
59 dep[dest] := dep[x] + 1;
60 tarjan(dest);
61 {if low[x]>=low[dest] then low[x] := low[dest]
62 else begin
63 inc(ans);
64 brid[dest] := true;
65 end; }
66 if low[dest]>dfn[x] then begin
67 inc(ans);
68 brid[dest] := true;
69 end;
70 if low[dest]<low[x] then low[x] := low[dest];
71 end
72 else begin
73 if visit[dest] then
74 if dfn[dest]<low[x] then low[x] := dfn[dest];
75 end;
76 i := next;
77 end;
78 visit[x] := false;
79 end;
80
81 procedure print;
82 begin
83 writeln(ans);
84 end;
85
86 procedure lca(x, y: longint);
87 var
88 i: longint;
89 begin
90 if dep[x]<dep[y] then begin
91 i := x; x := y; y := i;
92 end;
93 while dep[x]>dep[y] do begin
94 if brid[x] then begin
95 dec(ans);
96 brid[x] := false;
97 end;
98 x := pre[x];
99 end;
100 while x<>y do begin
101 if brid[x] then begin
102 dec(ans);
103 brid[x] := false;
104 end;
105 if brid[y] then begin
106 dec(ans);
107 brid[y] := false;
108 end;
109 x := pre[x]; y := pre[y];
110 end;
111 end;
112
113 procedure main;
114 var
115 i, x, y: longint;
116 begin
117 fillchar(visit, sizeof(visit), 0);
118 fillchar(pre, sizeof(pre), 0);
119 fillchar(dfn, sizeof(dfn), 0);
120 fillchar(dep, sizeof(dep), 0);
121 fillchar(brid, sizeof(brid), 0);
122 tarjan(1);
123 readln(q);
124 for i := 1 to q do begin
125 readln(x, y);
126 lca(x, y);
127 print;
128 end;
129 end;
130
131 begin
132 assign(input,'1.txt'); reset(input);
133 test := 0;
134 repeat
135 inc(test);
136 readln(n, m);
137 if n+m=0 then break;
138 writeln('Case ', test, ':');
139 init;
140 main;
141 writeln;
142 until false;
143 end.



原文地址:https://www.cnblogs.com/wmzisfoolish/p/2435199.html