9.树的遍历序转换
A. 已知前序中序求后序 ?
procedure Solve(pre,mid:string);
var i:integer;
begin
if (pre='') or (mid='') then exit;
i:=pos(pre[1],mid);
solve(copy(pre,2,i),copy(mid,1,i-1));
solve(copy(pre,i+1,length(pre)-i),copy(mid,i+1,length(mid)-i));
post:=post+pre[1]; {加上根，递归l束后post即ؓ后序遍历}
end;
B.已知中序后序求前序 ?
procedure Solve(mid,post:string);
var i:integer;
begin
if (mid='') or (post='') then exit;
i:=pos(post[length(post)],mid);
pre:=pre+post[length(post)]; {加上根，递归l束后pre即ؓ前序遍历}
solve(copy(mid,1,I-1),copy(post,1,I-1));
solve(copy(mid,I+1,length(mid)-I),copy(post,I,length(post)-i));
end;
C.已知前序后序求中序 ?
function ok(s1,s2:string):boolean;
var i,l:integer;
p:boolean;
begin
ok:=true;
l:=length(s1);
for i:=1 to l do
begin
p:=false;
for j:=1 to l do
if s1[i]=s2[j] then p:=true;
if not p then
begin
ok:=false;exit;
end;
end;
end;
procedure solve(pre,post:string);
var i:integer;
begin
if (pre='') or (post='') then exit;
i:=0;
repeat
inc(i);
until ok(copy(pre,2,i),copy(post,1,i));
solve(copy(pre,2,i),copy(post,1,i));
midstr:=midstr+pre[1];
solve(copy(pre,i+2,length(pre)-i-1),copy(post,i+1,length(post)-i-1));
end;
10.求图的弱q通子?DFS)
procedure dfs ( now,color: integer);
begin
for i:=1 to n do
if a[now,i] and c[i]=0 then
begin
c[i]:=color;
dfs(I,color);
end;
end;
12.q制转换
A.整数L正整数进刉的互化 ?
NOIP1996数制转换
讑֭W串A$的结构ؓ: A$='mp'
其中m为数字串(长度< =20),而n,p均ؓ1?位的数字?其中所表达的内容在2-10之间)
E序要求:从键盘上dA$?不用正确性检?,A$中的数字串m(nq制)以pq制的Ş式输?
例如:A$='48< 10 >8' 其意义ؓ:?0q制?8,转换?q制数输?
输出l果:48< 10 >=60< 8 >
B.实数L正整数进刉的互化 ?
C.负数q制Q ?NOIP2000 设计一个程序，d一个十q制数的基数和一个负q制数的基数Q?br />q将此十q制数{换ؓ此负 q制下的敎ͼ-R∈{-2Q?3Q?4,....-20}
13.全排列与l合的生成 ?排列的生成：Q?..nQ ?
procedure solve(dep:integer);
var i:integer;
begin
if dep=n+1 then
begin
writeln(s);
exit;
end;
for i:=1 to n do
if not used[i] then
begin
s:=s+chr(i+ord('0'));
used[i]:=true;
solve(dep+1);
s:=copy(s,1,length(s)-1);
used[i]:=false;
end;
end;
l合的生?1..n中选取k个数的所有方?
procedure solve(dep,pre:integer);
var i:integer;
begin
if dep=k+1 then
begin
writeln(s);
exit;
end;
for i:=1 to n do
if (not used[i]) and (i >pre) then
begin
s:=s+chr(i+ord('0'));
used[i]:=true;
solve(dep+1,i);
s:=copy(s,1,length(s)-1);
used[i]:=false;
end;
end;
14 递推关系 计算字串序号模型 USACO1.2.5 StringSobits
长度为N (N< =31)?1串中1的个数小于等于L的串l成的集合中扑և按大排序后的第I?1丌Ӏ?br />数字划分模型
*NOIP2001数的划分
整数n分成k份，且每份不能ؓI，
L两种分法不能相同(不考虑序)?
d[0,0]:=1;
for p:=1 to n do
for i:=p to n do
for j:=k downto 1 do inc(d[i,j],d[i-p,j-1]);
writeln(d[n,k]);
*变Ş1Q考虑序
d[ i, j] : = d [ i-k, j-1] (k=1..i)
*变Ş2Q若分解出来的每个数均有一个上限m
d[ i, j] : = d [ i-k, j-1] (k=1..m)
15.符优先法求解表辑ּ求值问?
const maxn=50;
var s1:array[1..maxn] of integer; {s1为数字栈}
s2:array[1..maxn] of char; {s2为算W栈}
t1,t2:integer; {栈顶指针}
procedure calcu;
var x1,x2,x:integer;
p:char;
begin
p:=s2[t2];
dec(t2);
x2:=s1[t1];
dec(t1);
x1:=s1[t1];
dec(t1);
case p of
'+':x:=x1+x2;
'-':x:=x1-x2;
'*':x:=x1*x2;
'/':x:=x1 div 2;
end;
inc(t1);
s1[t1]:=x;
end;
procedure work;
var c:char;
v:integer;
begin
t1:=0;
t2:=0;
read(c);
while c< >';' do
case c of
'+','-':
begin
while (t2 >0) and (s2[t2]< >'(') do calcu;
inc(t2);s2[t2]:=c;
read(c);
end ;
'*','/':
begin
if (t2 >0) and ((s2[t2]='*') or (s2[t2]='/')) then calcu;
inc(t2);s2[t2]:=c;
read(c);
end;
'(':
begin
inc(t2);
s2[t2]:=c;
read(c);
end;
')':
begin
while s2[t2]< >
'(' do calcu;
dec(t2);
read(c);
end;
'0'..'9':
begin
v:=0;
repeat
v:=10*v+ord(c)-ord('0');
read(c);
until (c< '0') or (c >'9');
inc(t1);
s1[t1]:=v;
end;
end;
while t2 >0 do calcu;
writeln(s1[t1]);
end;
16.查找法
折半查找
function binsearch(k:keytype):integer;
var low,hig,mid:integer;
begin
low:=1;
hig:=n;
mid:=(low+hig) div 2;
while (a[mid].key< >k) and (low< =hig) do
begin
if a[mid].key >k then hig:=mid-1
else low:=mid+1;
mid:=(low+hig) div 2;
end;
if low >hig then mid:=0;
binsearch:=mid;
end;
树Ş查找
?排序树：每个l点的值都大于其左子树Ml点的D小于其叛_树Q一l点的倹{ ?
查找
function treesrh(k:keytype):pointer;
var q:pointer;
begin
q:=root;
while (q< >nil) and (q^.key< >k) do if k<
q^.key then q:=q^.left else q:=q^.right;
treesrh:=q;
end;
]]>
1.数论法
求两数的最大公U数
function gcd(a,b:integer):integer;
begin
if b=0 then gcd:=a
else gcd:=gcd (b,a mod b);
end ;
求两数的最公倍数
function lcm(a,b:integer):integer;
begin
if a< b then swap(a,b);
lcm:=a;
while lcm mod b >
0 do inc(lcm,a);
end;
素数的求?
A.范围内判断一个数是否敎ͼ
function prime (n: integer): Boolean;
var I: integer;
begin
for I:=2 to trunc(sqrt(n)) do
if n mod I=0 then
begin
prime:=false; exit;
end;
prime:=true;
end;
B.判断longint范围内的数是否ؓ素数Q包含求50000以内的素数表Q：
procedure getprime;
var i,j:longint;
p:array[1..50000] of boolean;
begin
fillchar(p,sizeof(p),true);
p[1]:=false;
i:=2;
while i< 50000 do
begin
if p[i] then
begin
j:=i*2;
while j< 50000 do
begin
p[j]:=false;
inc(j,i);
end;
end;
inc(i);
end;
l:=0;
for i:=1 to 50000 do
if p[i] then
begin
inc(l);
pr[l]:=i;
end;
end;{getprime}
function prime(x:longint):integer;
var i:integer;
begin
prime:=false;
for i:=1 to l do
if pr[i] >=x then break
else if x mod pr[i]=0 then exit;
prime:=true;
end;{prime}
2.3.4.求最生成树
A.Prim法Q ?
procedure prim(v0:integer);
var lowcost,closest:array[1..maxn] of integer;
i,j,k,min:integer;
begin
for i:=1 to n do
begin
lowcost[i]:=cost[v0,i];
closest[i]:=v0;
end;
for i:=1 to n-1 do
begin {Lȝ成树最q的未加入顶点k}
min:=maxlongint;
for j:=1 to n do
if (lowcost[j]< min) and (lowcost[j]< >0) then
begin
min:=lowcost[j];
k:=j;
end;
lowcost[k]:=0; {顶点k加入生成树}
{生成树中增加一条新的边k到closest[k]}
{修正各点的lowcost和closest值}
for j:=1 to n do
if cost[k,j]< lwocost[j] then
begin
lowcost[j]:=cost[k,j];
closest[j]:=k;
end;
end;
end;{prim}
B.Kruskal法Q?贪心)
按权值递增序删去图中的边Q若不Ş成回路则此边加入最生成树?
function find(v:integer):integer; {q回点v所在的集合}
var i:integer;
begin
i:=1;
while (i< =n) and (not v in vset[i]) do inc(i);
if i< =n then find:=i else find:=0;
end;
procedure kruskal;
var tot,i,j:integer;
begin
for i:=1 to n do vset[i]:=[i];{初始化定义n个集合，WI个集合包含一个元素I}
p:=n-1; q:=1; tot:=0; {p为尚待加入的ҎQq集指针}
sort;
{Ҏ有边按权值递增排序Q存于e[i]中，e[i].v1与e[i].v2I所q接的两个顶点的序号Qe[i].len为第I条边的长度}
while p >0 do
begin
i:=find(e[q].v1);
j:=find(e[q].v2);
if i< >j then
begin
inc(tot,e[q].len);
vset[i]:=vset[i]+vset[j];
vset[j]:=[];
dec(p);
end;
inc(q);
end;
5.最短\径 ?
A.标号法求解单源点最短\径：
var a:array[1..maxn,1..maxn] of integer;
b:array[1..maxn] of integer; {b[i]指顶点i到源点的最短\径}
mark:array[1..maxn] of boolean;
procedure bhf;
var best,best_j:integer;
begin
fillchar(mark,sizeof(mark),false);
mark[1]:=true;
b[1]:=0;{1为源点}
repeat best:=0;
for i:=1 to n do
If mark[i] then {Ҏ一个已计算出最短\径的点}
for j:=1 to n do
if (not mark[j]) and (a[i,j] >0) then
if (best=0) or (b[i]+a[i,j]< best) then
begin
best:=b[i]+a[i,j]; best_j:=j;
end;
if best >0 then
begin
b[best_j]:=best;
mark[best_j]:=true;
end;
until best=0;
end;{bhf}
B.Floyed法求解所有顶点对之间的最短\径：
procedure floyed;
begin
for I:=1 to n do
for j:=1 to n do
if a[I,j] >0 then
p[I,j]:=I else p[I,j]:=0;
{p[I,j]表示I到j的最短\径上j的前q点}
for k:=1 to n do {枚D中间l点}
for i:=1 to n do for j:=1 to n do
if a[i,k]+a[j,k]< a[i,j] then
begin
a[i,j]:=a[i,k]+a[k,j];
p[I,j]:=p[k,j];
end;
end;
C. Dijkstra 法Q?
cM标号法，本质心算法?
var a:array[1..maxn,1..maxn] of integer;
b,pre:array[1..maxn] of integer; {pre[i]指最短\径上I的前q点}
mark:array[1..maxn] of boolean;
procedure dijkstra(v0:integer);
begin
fillchar(mark,sizeof(mark),false);
for i:=1 to n do
begin
d[i]:=a[v0,i];
if d[i]< >0 then
pre[i]:=v0
else
pre[i]:=0;
end;
mark[v0]:=true;
repeat {每@环一ơ加入一个离1集合最q的l点q调整其他结点的参数}
min:=maxint;
u:=0; {u记录?集合最q的l点}
for i:=1 to n do
if (not mark[i]) and (d[i]< min) then
begin
u:=i; min:=d[i];
end;
if u< >0 then
begin
mark[u]:=true;
for i:=1 to n do
if (not mark[i]) and (a[u,i]+d[u]< d[i]) then
begin
d[i]:=a[u,i]+d[u];
pre[i]:=u;
end;
end;
until u=0;
end;
D.计算囄传递闭?
Procedure Longlink;
Var T:array[1..maxn,1..maxn] of boolean;
Begin
Fillchar(t,sizeof(t),false);
For k:=1 to n do
For I:=1 to n do
For j:=1 to n do
T[I,j]:=t[I,j] or (t[I,k] and t[k,j]);
End;
7.排序法
A.快速排序：
procedure sort(l,r:integer);
var i,j,mid:integer;
begin
i:=l;j:=r;
mid:=a[(l+r) div 2];
{当前序列在中间位置的数定义Z间数}
repeat
while a[i]< mid do inc(i); {在左半部分寻找比中间数大的数}
while mid< a[j] do dec(j);{在右半部分寻找比中间数小的数}
if i< =j then
begin {若找Cl与排序目标不一致的数对则交换它们}
swap(a[i],a[j]);
inc(i);
dec(j); {l箋找}
end;
until i >j;
if l< j then
sort(l,j); {若未C个数的边界，则递归搜烦左右区间}
if i< r then sort(i,r);
end;{sort}
B.插入排序Q?
procedure insert_sort(k,m:word); {k为当前要插入的数Qm为插入位|的指针}
var i:word; p:0..1;
begin
p:=0;
for i:=m downto 1 do
if k=a[i] then exit;
repeat If k >a[m] then
begin
a[m+1]:=k; p:=1;
end
else
begin
a[m+1]:=a[m];
dec(m);
end;
until p=1;
end;{insert_sort}
l ȝ序中为：
a[0]:=0;
for I:=1 to n do insert_sort(b[i],I-1);
C.选择排序Q ?
procedure sort;
var i,j,k:integer;
begin
for i:=1 to n-1 do
begin
k:=i;
for j:=i+1 to n do
if a[j]< a[k] then
k:=j; {扑ևa[i]..a[n]中最的Ca[i]作交换}
if k< >i then
begin
a[0]:=a[k];
a[k]:=a[i];
a[i]:=a[0];
end;
end;
end;
D. 冒排序
procedure sort;
var i,j,k:integer;
begin
for i:=n downto 1 do
for j:=1 to i-1 do
if a[j] >a[i] then
begin
a[0]:=a[i];
a[i]:=a[j];
a[j]:=a[0];
end;
end;
E.堆排序：
procedure sift(i,m:integer);{调整以i为根的子树成为堆,m为结ҎL}
var k:integer;
begin
a[0]:=a[i];
k:=2*i;{在完全二*树中l点i的左孩子?*i,叛_子ؓ2*i+1}
while k< =m do
begin
if (k< m) and (a[k]< a[k+1]) then inc(k);{扑ևa[k]与a[k+1]中较大值}
if a[0]< a[k] then
begin
a[i]:=a[k];
i:=k;
k:=2*i;
end
else
k:=m+1;
end;
a[i]:=a[0]; {根攑֜合适的位置}
end;
procedure heapsort;
var j:integer;
begin
for j:=n div 2 downto 1 do sift(j,n);
for j:=n downto 2 do
begin
swap(a[1],a[j]);
sift(1,j-1);
end;
end;
F. 归ƈ排序
{a为序列表Qtmp助数l}
procedure merge(var a:listtype; p,q,r:integer);
{已排序好的子序列a[p..q]与a[q+1..r]合ƈ为有序的tmp[p..r]}
var I,j,t:integer;
tmp:listtype;
begin
t:=p;
i:=p;
j:=q+1;{t为tmp指针QI,j分别为左叛_序列的指针}
while (t< =r) do
begin
if (i< =q){左序列有剩余} and ((j >r) or (a[i]< =a[j])) then {满取左边序列当前元素的要求}
begin
tmp[t]:=a[i]; inc(i);
end
else
begin
tmp[t]:=a[j];
inc(j);
end;
inc(t);
end;
for i:=p to r do a[i]:=tmp[i];
end;{merge}
procedure merge_sort(var a:listtype; p,r: integer); {合ƈ排序a[p..r]}
var q:integer;
begin
if p< >r then
begin
q:=(p+r-1) div 2;
merge_sort (a,p,q);
merge_sort (a,q+1,r);
merge (a,p,q,r);
end;
end;
{main}
begin
merge_sort(a,1,n);
end.
writeln(tot);
end;
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