poj - 2553 (强连通分量)
题目来源： http://acm.pku.edu.cn/JudgeOnline/problem?id=2553
强连通分量，很简单的题目，照着模板水过了。
代码
1
#include
<
iostream
>
2
#include
<
algorithm
>
3
using
namespace
std;
4
5
const
int
M
=
10005
;
6
int
cnt, scnt, nn;
7
int
stack[M], ord[M], low[M], id[M], ans[M];
8
bool
instack[M], map[M];
9
struct
Node
10
{
11
int
to;
12
Node
*
next;
13
};
14
Node
*
g[M];
15
16
17
bool
cmp(
int
a,
int
b)
18
{
19
return
a
<
b;
20
}
21
22
void
Tarjan(
int
e)
//
Tarjan的DFS过程，不断更新 low 和 ord
23
{
24
low[e]
=
++
cnt;
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ord[e]
=
cnt;
26
stack[
++
stack[
0
]]
=
e;
27
instack[e]
=
true
;
28
Node
*
tmp
=
g[e];
29
30
while
(tmp
!=
NULL)
31
{
32
int
t
=
tmp
->
to;
33
if
(
!
ord[t])
34
{
35
Tarjan(t);
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if
(low[t]
<
low[e])
37
low[e]
=
low[t];
38
}
39
else
if
(instack[t]
&&
ord[t]
<
low[e])
40
low[e]
=
ord[t];
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tmp
=
tmp
->
next;
42
}
43
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if
(low[e]
==
ord[e])
//
找到强连通分量的根
45
{
46
int
t;
47
scnt
++
;
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do
//
找出强连通分支中的点
49
{
50
t
=
stack[stack[
0
]
--
];
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id[t]
=
scnt;
52
instack[t]
=
false
;
53
}
while
(t
!=
e);
54
}
55
56
return
;
57
}
58
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void
find_components(
int
n)
//
Tarjan主过程，依次找出没有DFS的点
60
{
61
memset(instack,
false
,
sizeof
(instack));
62
memset(ord,
0
,
sizeof
(ord));
63
cnt
=
scnt
=
0
;
64
stack[
0
]
=
0
;
//
记录栈里元素的个数
65
for
(
int
i
=
1
; i
<=
n; i
++
)
66
{
67
if
(
!
ord[i])
68
Tarjan(i);
69
}
70
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return
;
72
}
73
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void
insert(
int
u,
int
v)
75
{
76
Node
*
tmp
=
new
Node;
77
tmp
->
to
=
v;
78
tmp
->
next
=
g[u];
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g[u]
=
tmp;
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}
81
82
int
main()
83
{
84
int
n, m;
85
while
(scanf(
"
%d
"
,
&
n)
!=
EOF
&&
n)
86
{
87
scanf(
"
%d
"
,
&
m);
88
memset(g, NULL,
sizeof
(g));
89
for
(
int
i
=
0
; i
<
m; i
++
)
//
建图
90
{
91
int
u, v ;
92
scanf(
"
%d%d
"
,
&
u,
&
v);
93
insert(u, v);
94
}
95
96
find_components(n);
97
nn
=
scnt;
98
memset(map,
false
,
sizeof
(map));
99
100
for
(
int
i
=
1
; i
<=
n; i
++
)
101
{
102
Node
*
tmp;
103
tmp
=
g[i];
104
while
(tmp
!=
NULL)
105
{
106
int
e
=
tmp
->
to;
107
if
(id[e]
!=
id[i]
&&
!
map[id[i]])
108
map[id[i]]
=
true
;
109
tmp
=
tmp
->
next;
110
}
111
}
112
113
int
k
=
0
;
114
for
(
int
i
=
1
; i
<=
n; i
++
)
115
{
116
int
pos
=
id[i];
117
if
(
!
map[pos])
118
{
119
ans[k
++
]
=
i;
120
}
121
}
122
123
if
(k
>
0
)
124
{
125
sort(ans, ans
+
k, cmp);
126
printf(
"
%d
"
,ans[
0
]);
127
for
(
int
i
=
1
; i
<
k; i
++
)
128
printf(
"
%d
"
,ans[i]);
129
printf(
"
\n
"
);
130
}
131
else
132
printf(
"
\n\n
"
);
133
}
134
return
0
;
135
}
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137
138