157 条题解
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0
招财猪 LV 8 @ 2009-11-08 22:29:46
program v1143;
var
f:array[2..40,1..20,1..20,1..20]of longint;
d:array[1..20,1..20]of longint;
i,j,k,p,n,max,a,b,c,q,l:longint;
begin
read(n);
for i:=1 to n do
for j:=1 to n do read(d);
fillchar(f,sizeof(f),0);
f[2,1,1,1]:=d[1,1];
for p:=3 to n+1 do
for i:=1 to p-1 do
for j:=i to p-1 do
for k:=j to p-1 do
begin
max:=0;
for q:=0 to 7 do
begin
l:=q;
a:=i-(l mod 2);
l:=l shr 1;
b:=j-(l mod 2);
l:=l shr 1;
c:=k-l;
if (a>0)and(a0)and(b0)and(cmax then max:=f[p-1,a,b,c];
end;
f[p,i,j,k]:=max+d[p-i,i]+d[p-j,j]+d[p-k,k];
if (i=j)and(j=k) then
f[p,i,j,k]:=f[p,i,j,k]-2*d[p-i,i]
else if (i=j)or(i=k) then dec(f[p,i,j,k],d[p-i,i])
else if (j=k) then dec(f[p,i,j,k],d[p-j,j]);
end;
for p:=n+2 to 2*n do
for i:=1 to 2*n-p+1 do
for j:=i to 2*n-p+1 do
for k:=j to 2*n-p+1 do
begin
max:=0;
for q:=0 to 7 do
begin
l:=q;
a:=i+(l mod 2);
l:=l shr 1;
b:=j+(l mod 2);
l:=l shr 1;
c:=k+l;
if (a>0)and(a0)and(b0)and(cmax then max:=f[p-1,a,b,c];
end;
f[p,i,j,k]:=max+d[n-i+1,p-n+i-1]+d[n-j+1,p-n+j-1]+d[n-k+1,p-n+k-1];
if (i=j)and(j=k) then
f[p,i,j,k]:=f[p,i,j,k]-2*d[n-i+1,p-n+i-1]
else if (i=j)or(i=k) then dec(f[p,i,j,k],d[n-i+1,p-n+i-1])
else if (j=k) then dec(f[p,i,j,k],d[n-j+1,p-n+j-1]);
end;
writeln(f[2*n,1,1,1]);
end.
发现自己很菜鸟
弄了很久才出来 -
0
楼下俩 四维的怎么做的， 不太明白， 希望详细的讲解一下， 本人只会三维的！！！ d-rex 的程序， 很经典， 能加个注释吗？？
var
f : array[0..200,0..200,0..200]of longint;
map : array[0..200,0..200]of longint;
i, j, l, r, m, n, ans : longint;
function max(a,b,c,d,z,x,y,v:longint):longint;
begin
max := a;
if max < b then max := b;
if max < c then max := c;
if max < d then max := d;
if max < z then max := z;
if max < x then max := x;
if max < y then max := y;
if max < v then max := v;
end;
begin
readln(n);
for i := 1 to n do
for j := 1 to n do read(map);
fillchar(f,sizeof(f),0);
for i := 3 to n*2-3 do
for j := i-2 downto 1 do
for l := i-1 downto j+1 do
for r := i downto l+1 do
f[j,l,r] := max(f[j,l,r],
f[j,l-1,r-1],
f[j,l,r-1],
f[j,l-1,r],
f[j-1,l-1,r-1],
f[j-1,l,r],
f[j-1,l-1,r],
f[j-1,l,r-1])+map+map+map;
ans := f[n-2,n-1,n]+map[1,1]+map[1,2]+map[2,1]+map[n,n]+map[n,n-1]+map[n-1,n];
writeln(ans);
end. -
0
wangshizhou LV 8 @ 2009-11-08 18:06:53
var
f:array[0..30,0..30,0..30,0..50]of longint ;
a:array[0..20,0..20]of integer;
i,j,k,o,p,n,ans:longint;function max(a,b,c,d,e,f,g,h:longint):longint;
begin
max:=a;
if max -
0
var
f:array[-1..50,-1..30,-1..30,-1..30]of longint;
a:array[1..20,1..20]of longint;
i,j,l,k,t,n,m,temp,ans,max1:longint;
function max(a,b:longint):longint;
begin
max:=a;
if b>max then max:=b;
end;
begin
readln(n);
for i:=1 to n do
begin
for j:=1 to n do
read(a);
readln;
end;
f[1,1,1,1]:=a[1,1];
for k:=2 to 2*n-1 do
for i:=1 to n do
for j:=1 to n do
for t:=1 to n do
if (k-i+1>0)and(k-j+1>0)and(k-t+1>0) then
begin
temp:= max(max(max(f[k-1,i-1,j-1,t-1],f[k-1,i-1,j-1,t]),
max(f[k-1,i-1,j,t-1], f[k-1,i-1,j,t])),
max(max(f[k-1,i,j-1,t-1], f[k-1,i,j-1,t]),
max(f[k-1,i,j,t-1], f[k-1,i,j,t])));if (ij)and(jt)and(it) then temp:=temp+a+a[j,k-j+1]+a[t,k-t+1];
if (i=j) and(it) then temp:=temp+a+a[t,k-t+1];
if (i=t) and(ij) then temp:=temp+a+a[j,k-j+1];
if (j=t) and(ij) then temp:=temp+a[j,k-j+1]+a;
if (i=j) and(t=i) then temp:=temp+a;
if temp>f[k,i,j,t]
then f[k,i,j,t]:=temp;end;
write(f[2*n-1,n,n,n]);
end. -
0
动归，哦耶！！
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms -
0
const
maxn=20;
var
map:array[1..maxn,1..maxn] of longint;
n,k,i,ii,iii,j,jj,jjj,x,y,z,max:longint;
opt:array[0..maxn,0..maxn,0..maxn,0..3*maxn] of longint;
flag:boolean;
beginreadln(n);
for i:=1 to n do
begin
for j:=1 to n do
read(map[i][j]);
readln;
end;fillchar(opt,sizeof(opt),0);
opt[1][1][1][1]:=map[1][1];
for k:=2 to 2*n-1 do
for i:=1 to n do
for ii:=1 to n do
for iii:=1 to n do
begin
j:=k+1-i;
jj:=k+1-ii;
jjj:=k+1-iii;
flag:=false;
if (j>n) or (jn) or (jjn) or (jjj=max
then max:=opt[ii-y][iii-z][k-1];if (i=ii) and (ii=iii) then opt[i][ii][iii][k]:=max+map[i][j]
else if (i=ii) then opt[i][ii][iii][k]:=max+map[i][j]+map[iii][jjj]
else if (ii=iii) then opt[i][ii][iii][k]:=max+map[i][j]+map[ii][jj]
else if (i=iii) then opt[i][ii][iii][k]:=max+map[i][j]+map[ii][jj]
else opt[i][ii][iii][k]:=max+map[i][j]+map[ii][jj]+map[iii][jjj];
end;writeln(opt[n][n][n][2*n-1]);
end.
-
0
yjl3567618 LV 8 @ 2009-11-01 13:33:14
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0msvar i,j,k,m,l,q,n:longint;
a,b:array[0..21,0..21] of longint;
f:array[0..50,0..21,0..21,0..21] of longint;
function max(a,b:longint):longint;
begin
if a>b then max:=a
else max:=b;
end;
begin
readln(n);
for i:=1 to n do
begin
for j:=1 to n do
read(a);
readln;
end;
for i:=n+1 to 2*n-1 do
for j:=n downto n+1+i-2*n do
begin b:=a[j,i+1-j];end;
f[1,1,1,1]:=a[1,1];
for i:=2 to n do
for j:=1 to i do
for k:=j to i do
for l:=k to i do
begin
f:=max(f,f);
f:=max(f,max(f,f));
f:=max(f,max(f,f));
f:=max(f,max(f,f));
if (jK) and (kl) then f:=f+a+a+a;
if (j=k) and (kl) then f:=f+a+a;
if (l=k) and (jk) then f:=f+a+a;
if (j=k) and (k=l) then f:=f+a;
end;for i:=n+1 to 2*n-1 do
begin
q:=2*n-i;
for j:=1 to q do
for k:=j to q do
for l:=k to q do
begin
f:=max(f,f);
f:=max(f,max(f,f));
f:=max(f,max(f,f));
f:=max(f,max(f,f));
if (jk) and (kl) then f:=f+b+b+b;
if (j=k) and (kl) then f:=f+b+b;
if (l=k) and (jk) then f:=f+b+b;
if (j=k) and (k=l) then f:=f+b;end;
end;
writeln(f[2*n-1,1,1,1]);
end. -
0
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms
#include
using namespace std;
int a[21][21],b[41][21][21][21],c[8][3]={{0,0,0},{0,1,0},{0,0,1},{0,1,1},{1,0,0},{1,0,1},{1,1,0},{1,1,1}};
main()
{
int n,i,j,t,l,s,s1,k;
cin>>n;
for(i=1;ia[i][j];
for(i=2;i -
0
cz908640443 LV 7 @ 2009-10-30 12:55:23
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms
var
a:array[0..25,0..25]of longint;
f:array[1..40,-20..20,-20..20,-20..20]of longint;
i,j,k,n,x:longint;
procedure init;
begin
fillchar(f,sizeof(f),0);
readln(n);
for i:=1 to n do
for j:=1 to n do
read(a);
f[1,1,1,1]:=a[1,1];
end;
procedure try(p,q,r:longint);
begin
if f[x,i,j,k] -
0
var
a:array[0..40,0..40,0..40,0..40]of longint;
b:array[0..40,0..40]of longint;
c,d:array[1..100]of longint;
i,n,m,j,k,l,jl:longint;
function max(a1,a2,a3,a4,a5,a6,a7,a8:longint):longint;
begin
if (a1>=a2) and (a1>=a3) and (a1>=a4) and (a1>=a5) and (a1>=a6) and (a1>=a7)and (a1>=a8) then max:=a1
else if (a2>=a3) and (a2>=a4) and (a2>=a5) and (a2>=a6) and (a2>=a7) and (a2>=a8) then max:=a2
else if (a3>=a4) and (a3>=a5) and (a3>=a6) and (a3>=a7) and (a3>=a8) then max:=a3
else if (a4>=a5) and (a4>=a6) and (a4>=a7) and (a4>=a8) then max:=a4
else if (a5>=a6) and (a5>=a7) and (a5>=a8) then max:=a5
else if (a6>=a7) and (a6>=a8) then max:=a6
else if (a7>=a8) then max:=a7
else max:=a8;
end;
begin
readln(l);
for i:=1 to l do
for n:=1 to l do
read(b);
for i:=2 to l+1 do
begin
c[i]:=1;
d[i]:=d+1;
end;
for i:=l+2 to 2*l do
begin
c[i]:=c+1;
d[i]:=l;
end;
for i:=1 to l do
for n:=1 to l do
for m:=c to d do
for j:=c to d do
begin
jl:=b+b+b;
if (n=m) and (nj) then jl:=jl-b
else if (n=j) and (nm) then jl:=jl-b
else if (n=j) and (n=m) then jl:=b
else if (j=m) and (jn) then jl:=jl-b;
a:=max(a,a,a,a,a,a,a,a)+jl;
end;
writeln(a[l,l,l,l]);
end.水掉水掉...
简单万岁...
没优化万岁~
TAT
可不可以不要水——by糟糕胖 -
0
fan6882090 LV 7 @ 2009-10-27 21:56:59
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms
一次AC!
program fang;
var n,k,i,j,x,y,z,x1,y1,x2,y2,x3,y3,p:longint;
data:array[1..20,1..20] of longint;
ans:array[0..39,0..20,0..20,0..20] of longint;function max(a,b,c,d,e,f,g,h:longint):longint;
begin
if (a>=b) and (a>=c) and (a>=d) and (a>=e) and (a>=f) and (a>=g) and (a>=h)
then exit(a);
if (b>=a) and (b>=c) and (b>=d) and (b>=e) and (b>=f) and (b>=g) and (b>=h)
then exit(b);
if (c>=a) and (c>=b) and (c>=d) and (c>=e) and (c>=f) and (c>=g) and (c>=h)
then exit(c);
if (d>=a) and (d>=b) and (d>=c) and (d>=e) and (d>=f) and (d>=g) and (d>=h)
then exit(d);
if (e>=a) and (e>=b) and (e>=c) and (e>=d) and (e>=f) and (e>=g) and (e>=h)
then exit(e);
if (f>=a) and (f>=b) and (f>=c) and (f>=d) and (f>=e) and (f>=g) and (f>=h)
then exit(f);
if (g>=a) and (g>=b) and (g>=c) and (g>=d) and (g>=e) and (g>=f) and (g>=h)
then exit(g);
exit(h);
end;begin
readln(n);
for i:=1 to n do
begin
for j:=1 to n do
read(data);
readln;
end;
for k:=1 to 2*n-1 do
for i:=1 to k do
for j:=1 to k do
for p:=1 to k do
begin
if (i>n) or (j>n) or (p>n)
then break;
x1:=i;
x2:=j;
x3:=p;
y1:=k+1-i;
y2:=k+1-j;
y3:=k+1-p;
if (y1>n) or (y2>n) or (y3>n)
then continue;
ans[k,i,j,p]:=max(ans[k-1,i,j,p],ans[k-1,i-1,j,p],ans[k-1,i,j-1,p],ans[k-1,i,j,p-1],ans[k-1,i-1,j-1,p],ans[k-1,i-1,j,p-1],ans[k-1,i,j-1,p-1],ans[k-1,i-1,j-1,p-1])+data[x1,y1];
if x1x2
then ans[k,i,j,p]:=ans[k,i,j,p]+data[x2,y2];
if (x3x1) and (x3x2)
then ans[k,i,j,p]:=ans[k,i,j,p]+data[x3,y3];
end;
writeln(ans[2*n-1,n,n,n]);
end.
简明扼要! -
0
Program FangGeQuShu3;
Var i,j,k,l,m,n,tmp:Longint;
a1,a2,a3,a4:longint;
a:array[0..20,0..20] of longint;
f:array[0..40,0..20,0..20,0..20] of longint;
Function Max(x,y:longint):longint;
Begin
max:=x;
if y>max then max:=y;
End;
Begin
Readln(n);
For i:=1 to n do
for j:=1 to n do
read(a);
f[1,1,1,1]:=a[1,1];
for l:=2 to n*2-1 do
for i:=1 to n do
for j:=1 to n do
for k:=1 to n do
begin
if ((l-i+1>0) and (l-i+10) and (l-j+10) and (l-k+1f[l,i,j,k] then f[l,i,j,k]:=tmp;
end;
end;
writeln(f[n*2-1,n,n,n]);End.
-
0
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
---|---|---|---|---|---|---|---|-
Accepted 有效得分：100 有效耗时：0ms -
0
shenyanghua LV 8 @ 2009-10-23 18:16:54
-,-
为什么我死在开小数组?...改大就好了...
果然得先做二取..
按对角线分两部分。。其实不用这么麻烦var f:array[-10..40,-10..40,-10..40,-10..40] of longint;
a,b:array[1..1000,1..1000] of longint;
l:array[1..1000] of longint;
num1,num2,num3,num4,max1,max2,maxn,z1,z2,z3,x,y,n,m,i,j,k,ss:longint;
function max(x,y:longint):longint;
begin
if x>y then exit(x) else exit(y);
end;begin
readln(n);
for i:=1 to n do
for j:=1 to n do
read(b);
for i:=1 to n do
begin
x:=1;
y:=i;
l[i]:=1;a:=b[x,y];
while (x1) do
begin
inc(l[i]);
inc(x);
dec(y);
a:=b[x,y];
end;
end;for i:=n+1 to 2*n-1 do
begin
x:=i-n+1;
y:=n;
l[i]:=1;
a:=b[x,y];
while (x1) do
begin
inc(l[i]);
inc(x);
dec(y);
a:=b[x,y];
end;
end;for i:=1 to n do
for z1:=1 to l[i] do
for z2:=1 to l[i] do
for z3:=1 to l[i] do
begin
ss:=0;
if (z1z2) and (z1z3) and (z2z3) then
ss:=a+a+a;
if (z1z2) and (z2=z3) then
ss:=a+a;
if (z1z3) and (z2=z3) then
ss:=a+a;
if (z2z3) and (z1=z2) then
ss:=a+a;
if (z1=z2) and (z2=z3) then
ss:=a;
num1:=max(f,f);
num2:=max(f,f);
num3:=max(f,f);
num4:=max(f,f);
max1:=max(num1,num2);
max2:=max(num3,num4);
f:=max(max1,max2)+ss;
end;for i:=n+1 to 2*n-1 do
for z1:=1 to l[i] do
for z2:=1 to l[i] do
for z3:=1 to l[i] do
begin
ss:=0;
if (z1z2) and (z1z3) and (z2z3) then
ss:=a+a+a;
if (z1z2) and (z2=z3) then
ss:=a+a;
if (z1z3) and (z2=z3) then
ss:=a+a;
if (z2z3) and (z1=z2) then
ss:=a+a;
if (z1=z2) and (z2=z3) then
ss:=a;
num1:=max(f,f);
num2:=max(f,f);
num3:=max(f,f);
num4:=max(f,f);
max1:=max(num1,num2);
max2:=max(num3,num4);
f:=max(max1,max2)+ss;
end;writeln(f[2*n-1,1,1,1]);
end. -
0
边界条件成了大问题，一定要注意……继续动规……不做水人……
a:array[1..20,1..20]of integer;
f:array[0..40,0..20,0..20,0..20]of longint;
i,j,k,l,n,num,num1,num2,num3:longint;function max(x,y:longint):longint;
begin
if x>y then max:=x
else max:=y;
end;
function min(x,y:integer):integer;
begin
if x0)and(k+1-i0)and(k+1-j0)and(k+1-lf[k,i,j,l] then f[k,i,j,l]:=num;
end;
writeln(f[2*n-1,n,n,n]);
end. -
0
mintalfance LV 10 @ 2009-10-20 11:27:51
program p1143;
var a:array[0..20,0..20] of longint;
f:array[0..40,0..20,0..20,0..20] of longint;
i,j,k,l,m,n,min,r,x1,x2,x3,y1,y2,y3:longint;
function max(x,y:longint):longint;
begin if x1 then begin
if f[l-1,x1-1,x2-1,x3-1]>min then min:=f[l-1,x1-1,x2-1,x3-1];
if f[l-1,x1-1,x2-1,x3]>min then min:=f[l-1,x1-1,x2-1,x3];
if f[l-1,x1-1,x2,x3]>min then min:=f[l-1,x1-1,x2,x3];
if x3-x2>1 then
if f[l-1,x1-1,x2,x3-1]>min then min:=f[l-1,x1-1,x2,x3-1];
end;
if x3-x2>1 then
if f[l-1,x1,x2,x3-1]>min then min:=f[l-1,x1,x2,x3-1];
if x2-x1>1 then begin
if f[l-1,x1,x2-1,x3]>min then min:=f[l-1,x1,x2-1,x3];
if f[l-1,x1,x2-1,x3-1]>min then min:=f[l-1,x1,x2-1,x3-1];
end;
f[l,x1,x2,x3]:=min+a[x1,y1]+a[x2,y2]+a[x3,y3];
end;
min:=f[n*2-3,n-2,n-1,n]+a[n,n]+a[n-1,n]+a[n,n-1];
writeln(min);
end.八种状态
-
0
编译通过...
├ 测试数据 01：答案正确... 0ms
├ 测试数据 02：答案正确... 0ms
├ 测试数据 03：答案正确... 0ms
├ 测试数据 04：答案正确... 0ms
├ 测试数据 05：答案正确... 0ms
├ 测试数据 06：答案正确... 0ms
├ 测试数据 07：答案正确... 0ms
├ 测试数据 08：答案正确... 0ms
├ 测试数据 09：答案正确... 0ms
├ 测试数据 10：答案正确... 0ms
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Accepted 有效得分：100 有效耗时：0ms#include
int f[21][21][21][50],n,a[21][21];
int maxn(int a,int b)
{
return a>b?a:b;
}
int minn(int a,int b)
{
return a>b?b:a;
}
int plt(int i,int j,int k,int p)
{
int x=0;
x=maxn(f[i][j][k][p-1],x);
x=maxn(f[i][j][k-1][p-1],x);
x=maxn(f[i][j-1][k][p-1],x);
x=maxn(f[i][j-1][k-1][p-1],x);
x=maxn(f[j][k][p-1],x);
x=maxn(f[j][k-1][p-1],x);
x=maxn(f[j-1][k][p-1],x);
x=maxn(f[j-1][k-1][p-1],x);
int q,t,z;
q=p-i+1;
t=p-j+1;
z=p-k+1;
if (i==j&&j==k)
return x+a[i][q];
if (i==j)
return x+a[i][q]+a[k][z];
if (j==k)
return x+a[i][q]+a[j][t];
if (i==k)
return x+a[i][q]+a[j][t];
return x+a[i][q]+a[j][t]+a[k][z];
}
int main(void)
{
int i,j,k,p;
scanf("%d",&n);
for (i=1;i
招财猪
LV 8
@ 2009-11-08 22:29:46
