hdu 1385 Minimum Transport Cost
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1211 Accepted Submission(s): 304
本题并不是很难，关键是路径的选择，如果有正确的选择路径方法，对大家来说应该都不是太难。本题思想 最短路
代码：
代码
#include
<
stdio.h
>
#include
<
queue
>
#include
<
stack
>
using
namespace
std;
int
com(
const
void
*
a,
const
void
*
b)
{
return
*
(
int
*
)a
-*
(
int
*
)b;
};
typedef
struct
node
{
int
s,data;
bool
operator
<
(
const
node
&
a)
const
{
if
(a.s
==
s)
return
a.data
<
data;
return
a.s
<
s;
}
}NODE;
NODE cur,next;
int
pre[
1000
],dist[
1000
],start,end,n,a[
1000
][
1000
],b[
1000
];
stack
<
int
>
st;
int
test(
int
x,
int
y)
{
if
(pre[x]
==
pre[y]
&&
x
<
y)
return
1
;
if
(pre[x]
==
pre[y])
return
0
;
if
(pre[x]
==
start)
{
if
(test(x,pre[y]))
return
1
;
}
else
if
(pre[y]
==
start)
{
if
(test(pre[x],y))
return
1
;
}
else
{
if
(test(x,pre[y]))
return
1
;
if
(test(pre[x],y))
return
1
;
if
(test(pre[x],pre[y]))
return
1
;
}
return
0
;
}
void
dijkstra(
int
v0)
{
priority_queue
<
NODE
>
qu;
cur.data
=
v0;
cur.s
=
0
;
int
visit[
1000
],mark
=
0
,i;
memset(visit,
0
,
sizeof
(visit));
memset(dist,
-
1
,
sizeof
(dist));
qu.push (cur);
dist[v0]
=
0
;
while
(
!
qu.empty ())
{
cur
=
qu.top ();
while
(visit[cur.data])
{
qu.pop();
if
(qu.empty ())
{
mark
=
1
;
break
;
}
cur
=
qu.top();
}
if
(mark)
break
;
qu.pop();
if
(cur.data
==
end)
return
;
visit[cur.data]
=
1
;
for
(i
=
1
;i
<=
n;i
++
)
{
if
(
!
visit[i]
&&
a[cur.data][i]
!=-
1
)
{
if
(i
==
end)
{
if
(dist[i]
==-
1
||
dist[cur.data]
+
a[cur.data][i]
<=
dist[i])
{
if
(dist[i]
!=-
1
&&
dist[cur.data]
+
a[cur.data][i]
==
dist[i])
{
if
(test(cur.data,i))
pre[i]
=
cur.data;
}
else
{
pre[i]
=
cur.data;
}
next.data
=
i;
dist[i]
=
next.s
=
a[cur.data][i]
+
dist[cur.data];
qu.push(next);
}
}
else
{
if
(dist[i]
==-
1
||
dist[cur.data]
+
a[cur.data][i]
+
b[i]
<=
dist[i])
{
if
(dist[i]
!=-
1
&&
dist[cur.data]
+
a[cur.data][i]
+
b[i]
==
dist[i])
{
if
(test(cur.data,i))
pre[i]
=
cur.data;
}
else
{
pre[i]
=
cur.data;
}
next.data
=
i;
dist[i]
=
next.s
=
a[cur.data][i]
+
dist[cur.data]
+
b[i];
qu.push(next);
}
}
}
}
}
}
void
re(
int
x)
{
if
(x
==
start)
return
;
st.push(x);
re(pre[x]);
}
int
main()
{
//
freopen("q1456.in","r",stdin);
//
freopen("3.out","w",stdout);
int
i,j;
while
(scanf(
"
%d
"
,
&
n)
!=
EOF)
{
if
(n
==
0
)
break
;
for
(i
=
1
;i
<=
n;i
++
)
for
(j
=
1
;j
<=
n;j
++
)
scanf(
"
%d
"
,
&
a[i][j]);
for
(i
=
1
;i
<=
n;i
++
)
scanf(
"
%d
"
,
&
b[i]);
while
(
1
)
{
scanf(
"
%d%d
"
,
&
start,
&
end);
if
(start
==-
1
&&
end
==-
1
)
break
;
dijkstra(start);
printf(
"
From %d to %d :\n
"
,start,end);
re(end);
printf(
"
Path: %d
"
,start);
while
(
!
st.empty ())
{
printf(
"
-->%d
"
,st.top ());
st.pop();
}
printf(
"
\nTotal cost : %d\n
"
,dist[end]);
printf(
"
\n
"
);
}
}
return
0
;
}