hoj1142
最短路+dfs，dfs的时候要记忆化，并记得判断能否从该点来。
View Code
#include
<
iostream
>
#include
<
cstdio
>
#include
<
cstdlib
>
#include
<
cstring
>
using
namespace
std;
#define
INF 0x03F3F3F3F
const
int
N
=
1000
;
int
path[N], vis[N];
int
n, m;
int
cost[N][N], lowcost[N], dp[N];
void
input()
{
scanf(
"
%d
"
,
&
m);
memset(cost,
-
1
,
sizeof
(cost));
for
(
int
i
=
0
; i
<
m; i
++
)
{
int
a, b, c;
scanf(
"
%d%d%d
"
,
&
a,
&
b,
&
c);
a
--
;
b
--
;
cost[a][b]
=
c;
cost[b][a]
=
c;
}
}
void
Dijkstra(
int
beg)
{
int
i, j, min;
memset(vis,
0
,
sizeof
(vis));
vis[beg]
=
1
;
for
(i
=
0
; i
<
n; i
++
)
{
lowcost[i]
=
INF; path[i]
=
beg;
}
lowcost[beg]
=
0
;
path[beg]
=
-
1
;
//
树根的标记
int
pre
=
beg;
for
(i
=
1
; i
<
n; i
++
)
{
min
=
INF;
for
(j
=
0
; j
<
n; j
++
)
//
下面的加法可能导致溢出,INF不能取太大
if
(vis[j]
==
0
&&
cost[pre][j]
!=
-
1
&&
lowcost[pre]
+
cost[pre][j]
<
lowcost[j])
{
lowcost[j]
=
lowcost[pre]
+
cost[pre][j];
path[j]
=
pre;
}
for
(j
=
0
; j
<
n; j
++
)
if
(vis[j]
==
0
&&
lowcost[j]
<
min)
{
min
=
lowcost[j]; pre
=
j;
}
vis[pre]
=
1
;
//
cout << lowcost[pre] << endl;
}
}
int
dfs(
int
a)
{
if
(dp[a]
!=
-
1
)
return
dp[a];
int
ans
=
0
;
for
(
int
i
=
0
; i
<
n; i
++
)
if
(cost[i][a]
!=
-
1
&&
lowcost[i]
>
lowcost[a])
ans
+=
dfs(i);
dp[a]
=
ans;
return
dp[a];
}
int
main()
{
//
freopen("D:\\t.txt", "r", stdin);
while
(scanf(
"
%d
"
,
&
n)
!=
EOF
&&
n
!=
0
)
{
input();
Dijkstra(
1
);
memset(dp,
-
1
,
sizeof
(dp));
dp[
0
]
=
1
;
printf(
"
%d\n
"
, dfs(
1
));
}
return
0
;
}