poj 3268 Silver Cow Party & zoj 2008 Invitation Cards
参考 算法导论 p580 单终点最短路径问题
这两题是一样的求法，先正着求一次最短路，再将边反向，再求一次
矩阵表示图的话就是将矩阵转置一次，邻接表表示的话就是读入边的时候，建两个表
poj3268
dijkstra求最短路
#include
<
iostream
>
#include
<
string
.h
>
using
namespace
std;
const
int
MAXN
=
1001
;
int
n,m,x;
const
int
INF
=
0x7FFF
;
int
G[MAXN][MAXN];
int
dist1[MAXN];
int
sum[MAXN];
bool
used[MAXN];
void
in
(){
int
i,j,from,to,t;
cin
>>
n
>>
m
>>
x;
for
(i
=
1
; i
<=
n;i
++
){
for
(j
=
1
; j
<=
n; j
++
){
if
(i
!=
j){
G[i][j]
=
INF;
}
else
{
G[i][j]
=
0
;
}
}
}
for
(i
=
1
; i
<=
m;i
++
){
cin
>>
from
>>
to
>>
t;
G[from][to]
=
t;
}
memset(sum,
0
,
sizeof
(sum));
}
void
dijkstra(){
int
i,j;
for
(i
=
1
; i
<=
n;i
++
){
dist1[i]
=
G[x][i];
used[i]
=
false
;
}
used[x]
=
true
;
dist1[x]
=
0
;
for
(i
=
1
; i
<
n; i
++
){
int
min
=
INF,minIndex
=
x;
for
(j
=
1
; j
<=
n; j
++
)
if
(
!
used[j]
&&
dist1[j]
<
min){
min
=
dist1[j];
minIndex
=
j;
}
used[minIndex]
=
true
;
sum[minIndex]
+=
dist1[minIndex];
for
(j
=
1
; j
<=
n; j
++
){
if
(
!
used[j]
&&
dist1[j]
>
dist1[minIndex]
+
G[minIndex][j]){
dist1[j]
=
dist1[minIndex]
+
G[minIndex][j];
}
}
}
}
void
reverse(){
int
i,j,temp;
for
(i
=
1
; i
<=
n; i
++
){
for
(j
=
1
; j
<
i; j
++
){
temp
=
G[i][j];
G[i][j]
=
G[j][i];
G[j][i]
=
temp;
}
}
}
int
main(){
int
max
=
0
,i;
in
();
dijkstra();
reverse();
dijkstra();
for
(i
=
1
; i
<=
n; i
++
)
if
(max
<
sum[i])
max
=
sum[i];
cout
<<
max
<<
endl;
return
0
;
}
zoj2008
spfa 求最短路
#include
<
stdio.h
>
#include
<
queue
>
using
namespace
std;
const
int
MAXN
=
1000000
;
const
int
INF
=
0x3f3f3f3f
;
struct
edge{
int
u, v, w, next, next2;
}e[MAXN];
int
first[MAXN], first2[MAXN], dist[MAXN];
queue
<
int
>
que;
int
p, q, cnt;
bool
inq[MAXN];
void
spfa(){
dist[
1
]
=
0
;
inq[
1
]
=
true
;
while
(
!
que.empty())
que.pop();
que.push(
1
);
while
(
!
que.empty()){
int
x
=
que.front();
que.pop();
inq[x]
=
false
;
for
(
int
i
=
first[x]; i
!=
-
1
; i
=
e[i].next){
int
u
=
e[i].u, v
=
e[i].v, w
=
e[i].w;
if
(dist[v]
>
dist[u]
+
w){
dist[v]
=
dist[u]
+
w;
if
(
!
inq[v]){
inq[v]
=
true
;
que.push(v);
}
}
}
}
}
void
spfa2(){
for
(
int
i
=
1
; i
<=
p;
++
i){
inq[i]
=
false
;
dist[i]
=
INF;
}
dist[
1
]
=
0
;
inq[
1
]
=
true
;
while
(
!
que.empty())
que.pop();
que.push(
1
);
while
(
!
que.empty()){
int
x
=
que.front();
que.pop();
inq[x]
=
false
;
for
(
int
i
=
first2[x]; i
!=
-
1
; i
=
e[i].next2){
int
u
=
e[i].v, v
=
e[i].u, w
=
e[i].w;
if
(dist[v]
>
dist[u]
+
w){
dist[v]
=
dist[u]
+
w;
if
(
!
inq[v]){
inq[v]
=
true
;
que.push(v);
}
}
}
}
}
int
main(){
int
t;
scanf(
"
%d
"
,
&
t);
while
(t
--
){
scanf(
"
%d%d
"
,
&
p,
&
q);
for
(
int
i
=
1
; i
<=
p;
++
i){
first[i]
=
-
1
;
first2[i]
=
-
1
;
dist[i]
=
INF;
inq[i]
=
false
;
}
for
(
int
i
=
0
; i
<
q;
++
i){
scanf(
"
%d%d%d
"
,
&
e[i].u,
&
e[i].v,
&
e[i].w);
e[i].next
=
first[e[i].u];
first[e[i].u]
=
i;
e[i].next2
=
first2[e[i].v];
//
建立两个邻接表
first2[e[i].v]
=
i;
}
spfa();
int
ans
=
0
;
for
(
int
i
=
1
; i
<=
p;
++
i)
ans
+=
dist[i];
spfa2();
for
(
int
i
=
1
; i
<=
p;
++
i)
ans
+=
dist[i];
printf(
"
%d\n
"
,ans);
}
return
0
;
}