比较明显的网络流最小割模型,对于这种模型我们需要先求获利的和,然后减去代价即可。
我们对于第i个人来说, 如果选他,会耗费A[I]的代价,那么(source,i,a[i])代表选他之后的代价,如果不选他,我们会产生Σw[i][j] 1<=j<=n的代价,也就是这么多的利益我们无法得到,然后对于两个人的互相影响,连边(i,j,2*w[i][j]),代表如果不选i,选j的话,本来i中选j的利益得不到,又要损失j对i的影响为w[i][j]。所以共计损失2*w[i][j]。之后求最小割=最大流就行了。
/**************************************************************
Problem:
2039
User: BLADEVIL
Language: Pascal
Result: Accepted
Time:
860
ms
Memory:
47112
kb
****************************************************************/
//By BLADEVIL
var
n :
longint
;
pre, other, len :
array
[
0..4000010
]
of
longint
;
last :
array
[
0..1010
]
of
longint
;
l :
longint
;
source, sink :
longint
;
que, d :
array
[
0..1010
]
of
longint
;
ans :
longint
;
function
min(a,b:
longint
):
longint
;
begin
if
a>b
then
min:=b
else
min:=a;
end
;
procedure
connect(x,y,z:
longint
);
begin
inc(l);
pre[l]:=last[x];
last[x]:=l;
other[l]:=y;
len[l]:=z;
end
;
procedure
init;
var
i, j :
longint
;
x :
longint
;
sum :
longint
;
begin
read(n);
source:=n+
2
; sink:=source+
1
; l:=
1
;
for
i:=
1
to
n
do
begin
read(x);
connect(source,i,x);
connect(i,source,
0
);
end
;
for
i:=
1
to
n
do
begin
sum:=
0
;
for
j:=
1
to
n
do
begin
read(x);
if
x=
0
then
continue;
sum:=sum+x;
connect(i,j,x<<
1
);
connect(j,i,
0
);
ans:=ans+x;
end
;
connect(i,sink,sum);
connect(sink,i,
0
);
end
;
end
;
function
bfs:
boolean
;
var
h, t, cur :
longint
;
q, p :
longint
;
begin
fillchar(d,sizeof(d),
0
);
h:=
0
; t:=
1
;
que[
1
]:=source;
d[source]:=
1
;
while
h<t
do
begin
inc(h);
cur:=que[h];
q:=last[cur];
while
q<>
0
do
begin
p:=other[q];
if
(len[q]>
0
)
and
(d[p]=
0
)
then
begin
inc(t);
que[t]:=p;
d[p]:=d[cur]+
1
;
if
p=sink
then
exit(
true
);
end
;
q:=pre[q];
end
;
end
;
exit(
false
);
end
;
function
dinic(x,flow:
longint
):
longint
;
var
rest, tmp :
longint
;
q, p :
longint
;
begin
if
x=sink
then
exit(flow);
rest:=flow;
q:=last[x];
while
q<>
0
do
begin
p:=other[q];
if
(len[q]>
0
)
and
(d[p]=d[x]+
1
)
and
(rest>
0
)
then
begin
tmp:=dinic(p,min(len[q],rest));
dec(rest,tmp);
dec(len[q],tmp);
inc(len[q
xor
1
],tmp);
end
;
q:=pre[q];
end
;
exit(flow-rest);
end
;
procedure
main;
begin
while
bfs
do
ans:=ans-dinic(source,maxlongint
div
10
);
writeln
(ans);
end
;
begin
init;
main;
end
.