C C++ 算法实例大全

function gcd(a,b:integer):integer;begin if b=0 then gcd:=aelse gcd:=gcd (b,a mod b);end ; 

function lcm(a,b:integer):integer;beginif a<b then swap(a,b);lcm:=a;while lcm mod b>0 do inc(lcm,a);end; 

function prime (n: integer): Boolean;var I: integer;beginfor I:=2 to trunc(sqrt(n)) doif n mod I=0 then begin prime:=false; exit;end;prime:=true;end; 

procedure getprime;var i,j:longint;p:array[1..50000] of boolean;beginfillchar(p,sizeof(p),true);p[1]:=false;i:=2;while i<50000 do beginif p[i] then beginj:=i*2;while j<50000 do beginp[j]:=false;inc(j,i);end;end;inc(i);end;l:=0;for i:=1 to 50000 doif p[i] then begininc(l);pr[l]:=i;end;end;{getprime}function prime(x:longint):integer;var i:integer;beginprime:=false;for i:=1 to l doif pr[i]>=x then breakelse if x mod pr[i]=0 then exit;prime:=true;end;{prime} 

procedure prim(v0:integer);varlowcost,closest:array[1..maxn] of integer;i,j,k,min:integer;beginfor i:=1 to n do beginlowcost[i]:=cost[v0,i];closest[i]:=v0;end;for i:=1 to n-1 do begin{寻找离生成树最近的未加入顶点k}min:=maxlongint;for j:=1 to n doif (lowcost[j]<min) and (lowcost[j]<>0) then beginmin:=lowcost[j];k:=j;end;lowcost[k]:=0; {将顶点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 doif cost[k,j]<lwocost[j] then beginlowcost[j]:=cost[k,j];closest[j]:=k;end;end;end;{prim}

function find(v:integer):integer; {返回顶点v所在的集合}var i:integer;begini:=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;vartot,i,j:integer;beginfor i:=1 to n do vset[i]:=[i];{初始化定义n个集合，第I个集合包含一个元素I}p:=n-1; q:=1; tot:=0; {p为尚待加入的边数，q为边集指针}sort;{对所有边按权值递增排序，存于e[I]中，e[I].v1与e[I].v2为边I所连接的两个顶点的序号，e[I].len为第I条边的长度}while p>0 do begini:=find(e[q].v1);j:=find(e[q].v2);if i<>j then begininc(tot,e[q].len);vset[i]:=vset[i]+vset[j];vset[j]:=[];dec(p);end;inc(q);end;writeln(tot);end;

vara:array[1..maxn,1..maxn] of integer;b:array[1..maxn] of integer; {b[i]指顶点i到源点的最短路径}mark:array[1..maxn] of boolean; procedure bhf;varbest,best_j:integer;beginfillchar(mark,sizeof(mark),false);mark[1]:=true; b[1]:=0;{1为源点}repeatbest:=0;for i:=1 to n doIf mark[i] then {对每一个已计算出最短路径的点}for j:=1 to n doif (not mark[j]) and (a[i,j]>0) then if (best=0) or (b[i]+a[i,j]<best) then beginbest:=b[i]+a[i,j]; best_j:=j;end;if best>0 then beginb[best_j]:=best；mark[best_j]:=true;end;until best=0;end;{bhf}

procedure floyed;beginfor I:=1 to n dofor j:=1 to n doif a[I,j]>0 then p[I,j]:=I else p[I,j]:=0; {p[I,j]表示I到j的最短路径上j的前驱结点}for k:=1 to n do {枚举中间结点}for i:=1 to n dofor j:=1 to n doif a[i,k]+a[j,k]<a[i,j] then begina[i,j]:=a[i,k]+a[k,j];p[I,j]:=p[k,j];end;end; 

vara:array[1..maxn,1..maxn] of integer;b,pre:array[1..maxn] of integer; {pre[i]指最短路径上I的前驱结点}mark:array[1..maxn] of boolean;procedure dijkstra(v0:integer);beginfillchar(mark,sizeof(mark),false);for i:=1 to n do begind[i]:=a[v0,i];if d[i]<>0 then pre[i]:=v0 else pre[i]:=0;end;mark[v0]:=true;repeat {每循环一次加入一个离1集合最近的结点并调整其他结点的参数}min:=maxint; u:=0; {u记录离1集合最近的结点}for i:=1 to n doif (not mark[i]) and (d[i]<min) then beginu:=i; min:=d[i];end;if u<>0 then beginmark[u]:=true; for i:=1 to n doif (not mark[i]) and (a[u,i]+d[u]<d[i]) then begind[i]:=a[u,i]+d[u];pre[i]:=u;end;end;until u=0;end;

Procedure Longlink;VarT:array[1..maxn,1..maxn] of boolean;BeginFillchar(t,sizeof(t),false);For k:=1 to n doFor I:=1 to n doFor j:=1 to n do T[I,j]:=t[I,j] or (t[I,k] and t[k,j]);End; 

procedure dfs ( now,color: integer);beginfor i:=1 to n doif a[now,i] and c[i]=0 then begin {对结点I染色}c[i]:=color;dfs(I,color);end;end; 

x[I],y[I],t[I]分别表示第I条边的起点，终点和权。共n个结点和m条边。procedure bellman-fordbeginfor I:=0 to n-1 do d[I]:=+infinitive;d[0]:=0;for I:=1 to n-1 dofor j:=1 to m do {枚举每一条边}if d[x[j]]+t[j]<d[y[j]] then d[y[j]]:=d[x[j]]+t[j];for I:=1 to m doif d[x[j]]+t[j]<d[y[j]] then return false else return true;end; 

procedure search(k,v:integer); {搜索第k个物品，剩余空间为v}var i,j:integer;beginif v<best then best:=v;if v-(s[n]-s[k-1])>=best then exit; {s[n]为前n个物品的重量和}if k<=n then beginif v>w[k] then search(k+1,v-w[k]);search(k+1,v);end;end; l DPF[I,j]为前i个物品中选择若干个放入使其体积正好为j的标志，为布尔型。实现:将最优化问题转化为判定性问题f [I, j] = f [ i-1, j-w[i] ] (w[I]<=j<=v) 边界：f[0,0]:=true.For I:=1 to n doFor j:=w[I] to v do F[I,j]:=f[I-1,j-w[I]];优化：当前状态只与前一阶段状态有关，可降至一维。F[0]:=true;For I:=1 to n do beginF1:=f;For j:=w[I] to v doIf f[j-w[I]] then f1[j]:=true;F:=f1;End;

Procedure update;var j,k:integer;beginc:=a;for j:=0 to n doif a[j]>0 thenif j+now<=n then inc(c[j+now],a[j]);a:=c;end; 

BeginFillChar(f,SizeOf(f),0);For i:=1 To M DoFor j:=1 To N DoIf i-problem[j].time>=0 ThenBegint:=problem[j].point+f[i-problem[j].time];If t>f[i] Then f[i]:=t;End;Writeln(f[M]);End. 

procedure try(dep:integer);var i,j:integer;begincal; {此过程计算当前系数的计算结果，now为结果}if now>n then exit; {剪枝}if dep=l+1 then begin {生成所有系数}cal;if now=n then inc(tot);exit;end;for i:=0 to n div pr[dep] do beginxs[dep]:=i;try(dep+1);xs[dep]:=0;end;end; 

procedure try(dep,rest:integer);var i,j,x:integer;beginif (rest<=0) or (dep=l+1) then beginif rest=0 then inc(tot);exit;end;for i:=0 to rest div pr[dep] dotry(dep+1,rest-pr[dep]*i);end;{main: try(1,n); } 

Procedure update;var j,k:integer;beginc:=a;for j:=0 to n doif a[j]>0 thenfor k:=1 to n div now doif j+now*k<=n then inc(c[j+now*k],a[j]);a:=c;end;{main}begin read(now); {读入第一个物品的重量}i:=0; {a[i]为背包容量为i时的放法总数}while i<=n do begin a[i]:=1; inc(i,now); end; {定义第一个物品重的整数倍的重量a值为1，作为初值}for i:=2 to v dobeginread(now);update; {动态更新}end;writeln(a[n]);

procedure qsort(l,r:integer);var i,j,mid:integer;begini:=l;j:=r; mid:=a[(l+r) div 2]; {将当前序列在中间位置的数定义为中间数}repeatwhile a[i]<mid do inc(i); {在左半部分寻找比中间数大的数}while a[j]>mid do dec(j);{在右半部分寻找比中间数小的数}if i<=j then begin {若找到一组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); {继续找}end;until i>j;if l<j then qsort(l,j); {若未到两个数的边界，则递归搜索左右区间}if i<r then qsort(i,r);end;{sort} 

procedure insert_sort;var i,j:integer;beginfor i:=2 to n do begina[0]:=a[i];j:=i-1;while a[0]<a[j] do begina[j+1]:=a[j];j:=j-1;end;a[j+1]:=a[0];end;end;{inset_sort} 

procedure sort;var i,j,k:integer;beginfor i:=1 to n-1 do for j:=i+1 to n doif a[i]>a[j] then swap(a[i],a[j]);end; 

procedure bubble_sort;var i,j,k:integer;beginfor i:=1 to n-1 dofor j:=n downto i+1 doif a[j]<a[j-1] then swap( a[j],a[j-1]); {每次比较相邻元素的关系}end; 

procedure sift(i,m:integer);{调整以i为根的子树成为堆,m为结点总数}var k:integer;begina[0]:=a[i]; k:=2*i;{在完全二叉树中结点i的左孩子为2*i,右孩子为2*i+1}while k<=m do beginif (k<m) and (a[k]<a[k+1]) then inc(k);{找出a[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; endelse k:=m+1;end;a[i]:=a[0]; {将根放在合适的位置}end; procedure heapsort;varj:integer;beginfor j:=n div 2 downto 1 do sift(j,n);for j:=n downto 2 do beginswap(a[1],a[j]);sift(1,j-1);end;end; 

var I,j,t:integer;tmp:listtype;begint:=p;i:=p;j:=q+1;{t为tmp指针，I,j分别为左右子序列的指针}while (t<=r) do beginif (i<=q){左序列有剩余} and ((j>r) or (a[i]<=a[j])) {满足取左边序列当前元素的要求}then begintmp[t]:=a[i]; inc(i);endelse begintmp[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); {合并排序a[p..r]}var q:integer;beginif p<>r then beginq:=(p+r-1) div 2;merge_sort (a,p,q);merge_sort (a,q+1,r);merge (a,p,q,r);end;end;{main}beginmerge_sort(a,1,n);end. 

procedure plus ( a,b:hp; var c:hp);var i,len:integer;beginfillchar(c,sizeof(c),0);if a[0]>b[0] then len:=a[0] else len:=b[0];for i:=1 to len do begininc(c[i],a[i]+b[i]);if c[i]>10 then begin dec(c[i],10); inc(c[i+1]); end; {进位}end;if c[len+1]>0 then inc(len);c[0]:=len;end;{plus} 

procedure substract(a,b:hp;var c:hp); var i,len:integer;beginfillchar(c,sizeof(c),0);if a[0]>b[0] then len:=a[0] else len:=b[0];for i:=1 to len do begininc(c[i],a[i]-b[i]);if c[i]<0 then begin inc(c[i],10);dec(c[i+1]); end;while (len>1) and (c[len]=0) do dec(len);c[0]:=len;end; 

procedure multiply(a:hp;b:longint;var c:hp);var i,len:integer;beginfillchar(c,sizeof(c),0);len:=a[0];for i:=1 to len do begininc(c[i],a[i]*b);inc(c[i+1],(a[i]*b) div 10);c[i]:=c[i] mod 10;end;inc(len);while (c[len]>=10) do begin {处理最高位的进位}c[len+1]:=c[len] div 10;c[len]:=c[len] mod 10;inc(len);end;while (len>1) and (c[len]=0) do dec(len); {若不需进位则调整len}c[0]:=len;end;{multiply} 

procedure high_multiply(a,b:hp; var c:hp}var i,j,len:integer;beginfillchar(c,sizeof(c),0);for i:=1 to a[0] dofor j:=1 to b[0] do begininc(c[i+j-1],a[i]*b[j]);inc(c[i+j],c[i+j-1] div 10);c[i+j-1]:=c[i+j-1] mod 10;end;len:=a[0]+b[0]+1;while (len>1) and (c[len]=0) do dec(len);c[0]:=len;end; 

procedure devide(a:hp;b:longint; var c:hp; var d:longint);{c:=a div b; d:= a mod b}var i,len:integer;beginfillchar(c,sizeof(c),0);len:=a[0]; d:=0;for i:=len downto 1 do begind:=d*10+a[i];c[i]:=d div b;d:=d mod b;end;while (len>1) and (c[len]=0) then dec(len);c[0]:=len;end; 

procedure high_devide(a,b:hp; var c,d:hp);vari,len:integer;beginfillchar(c,sizeof(c),0);fillchar(d,sizeof(d),0);len:=a[0];d[0]:=1;for i:=len downto 1 do beginmultiply(d,10,d);d[1]:=a[i];while(compare(d,b)>=0) do {即d>=b}beginSubtract(d,b,d);inc(c[i]);end;end;while(len>1)and(c.s[len]=0) do dec(len);c.len:=len;end; 

procedure Solve(pre,mid:string);var i:integer;beginif (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，递归结束后post即为后序遍历}end; 

procedure Solve(mid,post:string);var i:integer;beginif (mid='''') or (post='''') then exit;i:=pos(post[length(post)],mid);pre:=pre+post[length(post)]; {加上根，递归结束后pre即为前序遍历}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; 

function ok(s1,s2:string):boolean;var i,l:integer; p:boolean;beginok:=true;l:=length(s1);for i:=1 to l do beginp:=false;for j:=1 to l doif 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;beginif (pre='''') or (post='''') then exit;i:=0;repeatinc(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; 

procedure solve(dep:integer);vari:integer;beginif dep=n+1 then begin writeln(s);exit; end;for i:=1 to n doif not used[i] then begins:=s+chr(i+ord(''0''));used[i]:=true;solve(dep+1);s:=copy(s,1,length(s)-1); used[i]:=false;end;end; 

procedure solve(dep,pre:integer);vari:integer;beginif dep=k+1 then begin writeln(s);exit; end;for i:=1 to n doif (not used[i]) and (i>pre) then begins:=s+chr(i+ord(''0''));used[i]:=true;solve(dep+1,i);s:=copy(s,1,length(s)-1); used[i]:=false;end;end; 

function binsearch(k:keytype):integer;var low,hig,mid:integer;beginlow:=1;hig:=n;mid:=(low+hig) div 2;while (a[mid].key<>k) and (low<=hig) do beginif a[mid].key>k then hig:=mid-1else low:=mid+1;mid:=(low+hig) div 2;end;if low>hig then mid:=0;binsearch:=mid;end; 

function treesrh(k:keytype):pointer;var q:pointer;beginq:=root;while (q<>nil) and (q^.key<>k) doif k<q^.key then q:=q^.leftelse q:=q^.right;treesrh:=q;end; 

procedure try(i:byte);var j:byte;beginif i=n+1 then begin print;exit;end;for j:=1 to n doif a[i] and b[j+i] and c[j-i] then beginx[i]:=j;a[j]:=false; b[j+i]:=false; c[j-i]:=false;try(i+1);a[j]:=true; b[i+j]:=true; c[j-i]:=true;end;end;

h(n)=2*h(n-1)+1 h(1)=1初始所有铜片都在a柱上procedure hanoi(n,a,b,c:byte); {将第n块铜片从a柱通过b柱移到c柱上}beginif n=0 then exit;hanoi(n-1,a,c,b); {将上面的n-1块从a柱通过c柱移到b柱上}write(n,'moved from',a,'to',c);hanoi(n-1,b,a,c);{ 将b上的n-1块从b柱通过a柱移到c柱上end; 

Procedure move(k,goal:integer); {将最大不到位的k移到目标柱goal上}BeginIf k=0 then exit;For I:=1 to 3 doFor j:=1 to han[I,0] doIf h[I,j]=k then begin now:=I;nowp:=j; end; {找到k的位置}If now<>goal then begin {若未移到目标}Move(k-1,6-now-goal); {剩下的先移到没用的柱上}Writeln(k moved from now to goal);H[goal,h[goal,0]+1]:=h[now,nowp]; h[now,nowp]:=0;Inc(h[goal,0]); dec(h[now,0]);Move(k-1,goal); {剩下的移到目标上}End;

procedure work(dep,pre,s:longint); {入口为work(1,1,n)}{dep为当前试放的第dep个数,pre为前一次试放的数,s为当前剩余可分的总数}var j:longint;begin if dep=n then begin if s>=pre then inc(r); exit; end; for j:=pre to s div 2 do work(dep+1,j,s-j); end;类似：procedure try(dep:integer);var i:integer;beginif dep=k then beginif tot>=a[dep-1] then inc(sum);exit; end;for i:=a[dep-1] to tot div 2 do begina[dep]:=i; dec(tot,i); try(dep+1);inc(tot,i);end;end;{try} 

head:=1; tail:=0;while tail<head do begininc(tail);for k:=1 to n doif k方向可扩展 then begininc(head);list[head].x:=list[tail].x+dx[k]; {扩展出新结点list[head]}list[head].y:=list[tail].y+dy[k];处理新结点list[head];end;end; 

loc(I:integer):pointer; {寻找链表中的第I个结点的指针}procedure loc(L:linklist; I:integer):pointer;var p:pointer;j:integer;beginp:=L.head; j:=0;if (I>=1) and (I<=L.len) then while j<I do begin p:=p^.next; inc(j); end;loc:=p;end; 

procedure insert(L:linklist; I:integer; x:datatype);var p,q:pointer;beginp:=loc(L,I);new(q);q^.data:=x;q^.next:=p^.next;p^.next:=q;inc(L.len);end; 

procedure delete(L:linklist; I:integer);var p,q:pointer;beginp:=loc(L,I-1);q:=p^.next;p^.next:=q^.next;dispose(q);dec(L.len);end; 

p:=loc(L,I);new(q);q^.data:=x;q^.pre:=p;q^.next:=p^.next;p^.next:=q;q^.next^.pre:=q; 

p:=loc(L,I); {p为要删除的结点}p^.pre^.next:=p^.next;p^.next^.pre:=p^.pre;dispose(p);