POJ 2197 Jill's Tour Paths (DFS)
题意：给定n个城市及其之间的距离，以及距离限制limit 初始点start 结束点end，要求求出从start->end的所有不大于limit的路径并输出，而且是按字典序输出。
思路：很明显DFS+记录路径，由于要求字典序输出，那么建立前向星表的时候就要注意了，搜索子节点的顺序一定是从小到大的！另外就是输出格式上注意一下，格式很huang很baoli！
#include
<
iostream
>
#include
<
cstdio
>
#include
<
algorithm
>
#include
<
memory.h
>
#include
<
cmath
>
#include
<
bitset
>
#include
<
queue
>
#include
<
vector
>
using
namespace
std;
const
int
BORDER
=
(
1
<<
20
)
-
1
;
const
int
MAXSIZE
=
37
;
const
int
MAXN
=
30
;
const
int
INF
=
0x4ffffff
;
#define
CLR(x,y) memset(x,y,sizeof(x))
#define
ADD(x) x=((x+1)&BORDER)
#define
IN(x) scanf("%d",&x)
#define
OUT(x) printf("%d\n",x)
#define
MIN(m,v) (m)<(v)?(m):(v)
#define
MAX(m,v) (m)>(v)?(m):(v)
#define
ABS(x) ((x)>0?(x):-(x))
int
n,m,ans;
int
start,end,limit;
int
index,net[MAXN];
int
g[MAXN][MAXN];
bool
visit[MAXN];
int
route[MAXN];
int
deep;
int
case_id;
typedef
struct
{
int
v,val,next;
}EDGE;
typedef
struct
{
int
len,num;
int
city[MAXN];
}NODE;
EDGE edge[MAXN
*
MAXN];
NODE node[MAXN
*
MAXN
*
MAXN
*
MAXN];
int
init()
{
CLR(net,
-
1
);
CLR(visit,
0
);
CLR(g,
0
);
index
=
0
;
ans
=
0
;
deep
=
0
;
return
0
;
}
void
add_edge(
const
int
&
u,
const
int
&
v,
const
int
&
val)
{
edge[index].next
=
net[u];
edge[index].v
=
v;
edge[index].val
=
val;
net[u]
=
index;
++
index;
}
bool
cmp(
const
NODE
&
a,
const
NODE
&
b)
{
return
a.len
<
b.len;
}
/*
{
if(a.len != b.len)
return a.len < b.len;
int t = MAX(a.num,b.num);
for(int i = 1; i <= t; ++i)
if(a.city[i] != b.city[i])
return a.city[i] < b.city[i];
return true;
}
*/
int
input()
{
int
i,j,tmp;
int
a,b;
scanf(
"
%d
"
,
&
m);
for
(i
=
1
; i
<=
m;
++
i)
{
scanf(
"
%d%d%d
"
,
&
a,
&
b,
&
tmp);
g[a][b]
=
g[b][a]
=
tmp;
}
scanf(
"
%d%d
"
,
&
start,
&
end);
scanf(
"
%d
"
,
&
limit);
return
0
;
}
int
make_graph()
{
int
i,j,tmp;
for
(i
=
1
; i
<=
n;
++
i)
for
(j
=
n; j
>=
1
;
--
j)
if
(g[i][j])
add_edge(i,j,g[i][j]);
return
0
;
}
void
dfs(
const
int
&
u,
const
int
&
dist)
{
if
(dist
>
limit)
return
;
route[deep]
=
u;
++
deep;
if
(u
==
end)
{
memcpy(node[ans].city,route,deep
*
sizeof
(route[
0
]));
node[ans].len
=
dist;
node[ans].num
=
deep;
++
ans;
--
deep;
return
;
}
for
(
int
i
=
net[u] ; i
!=
-
1
; i
=
edge[i].next)
if
(
!
visit[edge[i].v])
{
visit[edge[i].v]
=
true
;
dfs(edge[i].v,dist
+
edge[i].val);
visit[edge[i].v]
=
false
;
}
--
deep;
}
int
output()
{
int
i,j,tmp;
printf(
"
Case %d:\n
"
,case_id);
if
(ans
==
0
)
{
printf(
"
NO ACCEPTABLE TOURS\n\n
"
);
return
0
;
}
for
(i
=
0
; i
<
ans;
++
i)
{
printf(
"
%d:
"
,node[i].len);
for
(j
=
0
; j
<
node[i].num;
++
j)
printf(
"
%d
"
,node[i].city[j]);
printf(
"
\n
"
);
}
printf(
"
\n
"
);
return
0
;
}
int
work()
{
make_graph();
visit[start]
=
true
;
dfs(start,
0
);
sort(node,node
+
ans,cmp);
return
0
;
}
int
main()
{
case_id
=
1
;
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
{
if
(n
==
-
1
)
break
;
init();
input();
work();
output();
++
case_id;
}
return
0
;
}