拆點即可。
在乙個寬m，長n的矩陣中，請你編乙個程式，n次從矩陣的左上角走到矩陣的右下角，每到一處，就取走該處的數字，請你選擇一種走法使取得的數字的和最大,並輸出其最大值。其中：3<=m<=20 m<=n<=100 1<=n<=10
測試資料 #1: accepted, time=15ms, mem=1154kb, score=10
測試資料 #2: accepted, time=0ms, mem=1154kb, score=10
測試資料 #3: accepted, time=15ms, mem=1154kb, score=10
測試資料 #4: accepted, time=0ms, mem=1154kb, score=10
測試資料 #5: accepted, time=15ms, mem=1154kb, score=10
測試資料 #6: accepted, time=0ms, mem=1154kb, score=10
測試資料 #7: accepted, time=0ms, mem=1154kb, score=10
測試資料 #8: accepted, time=0ms, mem=1158kb, score=10
測試資料 #9: accepted, time=0ms, mem=1150kb, score=10
測試資料 #10: accepted, time=0ms, mem=1150kb, score=10
time = 45ms mem = 1158kb score= 100
program p1413;const
inf=10000000;
type
rec=record
s,e,c,w,flow,next:longint;
end;
var v:array[0..4002] of boolean;
a:array[0..8002] of rec;
b,q,d,cur:array[0..4002] of longint;
n,m,k,i,j,st,tar,va,top:longint;
function poi2num(x,y:longint;isin:boolean):longint;inline;//num 1~2mn
begin
exit((x-1)*m*2+(y-1)*2+2-ord(isin));
end;
function min(a,b:longint):longint;inline;begin if aopen do
begin
close:=close mod n+1;
x:=q[close];
u:=b[x];
while u<>b[0] do
begin
if (d[a[u].e]st do
begin
delta:=min(delta,a[cur[u]].w-a[cur[u]].flow);
u:=y;
y:=a[cur[y]].s;
end;
u:=tar;
y:=a[cur[u]].s;
while u<>st do
begin
inc(a[cur[u]].flow,delta);
inc(addflow,a[cur[u]].c*delta);
dec(a[cur[u] xor 1].flow,delta);
u:=y;
y:=a[cur[y]].s;
end;
end;
function costflow:longint;
begin
costflow:=0;
while spfa do
inc(costflow,addflow);
end;
begin
fillchar(b,sizeof(b),$ff);
readln(k,m,n); top:=-1;
for i:=1 to n do
begin
for j:=1 to m do
begin
read(va);
add(poi2num(i,j,true),poi2num(i,j,false),1,va);
if (i=n) and (j=m) then add(poi2num(n,m,true),poi2num(n,m,false),k-1,0) else
if (i=1) and (j=1) then add(poi2num(1,1,true),poi2num(1,1,false),k-1,0) else
add(poi2num(i,j,true),poi2num(i,j,false),inf,0);
if iif jend;
readln;
end;
st:=1;tar:=2*m*n;
n:=2*m*n;
writeln(costflow);
end.
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