数据结构之顺序栈SqStack

顺序栈SqStack

基本操作

Status InitStack()//构造一个空栈SStatus DestroyStack()//销毁栈S，S不再存在Status ClearStack()//把S置为空栈Status StackEmpty()//若S为空栈，则返回true，否则返回falseint StackLength()//返回S的元素个数，即栈的长度Status GetTop(SElemType &e)//若栈不空，则用e返回S的栈顶元素，并返回OK，否则返回ERRORStatus Push(SElemType e)//插入元素e为新的栈顶元素Status Pop(SElemType &e)//若栈不空，则删除S的栈顶元素，并用e返回其值，并返回OK，否则返回ERRORStatus StackTraverse(int p)//从栈底到栈顶依次对栈中每个元素调用函数visit().一旦visit()失败则操作失败

几个小程序（代码正误检验，数制转换，括号匹配检验，行编辑程序）

CheckStackCode();//测试Stack代码正确性Conversion();//数制转换BracketsMatching();//括号匹配的检验LineEdit();//行编辑程序

////by coolxxx//#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<stack>#include<queue>#include<set>#include<bitset>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>//#include<stdbool.h>#include<math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-10)#define J 10000#define mod 1000000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#pragma comment(linker,"/STACK:1024000000,1024000000")#define N 104using namespace std;typedef long long LL;/*double anss;LL aans,sum;int cas,cass;int n,m,lll,ans;*/const int OK=1;const int ERROR=0;const int INFEASIBLE=-1;const int STACK_INIT_SIZE=100;//存储空间初始分配量const int STACKINCREMENT=10;//存储空间分配增量typedef int Status;typedef int SElemType;Status OutputInt(SElemType e);Status OutputChar(SElemType e);typedef struct{SElemType *base;//栈构造之前和销毁之后，base的值为NULLSElemType *top;//栈顶指针int stacksize;//当前已分配的存储空间，以元素为单位Status InitStack()//构造一个空栈S{base=(SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));if(!base)exit(OVERFLOW);//存储分配失败top=base;stacksize=STACK_INIT_SIZE;return OK;}//InitStackStatus DestroyStack()//销毁栈S，S不再存在{free(base);base=NULL;top=NULL;stacksize=0;return OK;}//DestroyStackStatus ClearStack()//把S置为空栈{top=base;return OK;}//ClearStackStatus StackEmpty()//若S为空栈，则返回true，否则返回false{if(top==base)return true;return false;}//StackEmptyint StackLength()//返回S的元素个数，即栈的长度{return top-base;}//StackLengthStatus GetTop(SElemType &e)//若栈不空，则用e返回S的栈顶元素，并返回OK，否则返回ERROR{if(top==base)return ERROR;e=*(top-1);return OK;}//GetTopStatus Push(SElemType e)//插入元素e为新的栈顶元素{if(top-base>=stacksize)//栈满，追加存储空间{base=(SElemType *)realloc(base,(stacksize+STACKINCREMENT)*sizeof(SElemType));if(!base)exit(OVERFLOW);//存储分配失败top=base+stacksize;stacksize+=STACKINCREMENT;}(*top++)=e;return OK;}//PushStatus Pop(SElemType &e)//若栈不空，则删除S的栈顶元素，并用e返回其值，并返回OK，否则返回ERROR{if(top==base)return ERROR;e=(*--top);return OK;}//PopStatus StackTraverse(int p)//从栈底到栈顶依次对栈中每个元素调用函数visit().一旦visit()失败则操作失败{SElemType *i=base;Status (*visit)(SElemType);if(p==1)visit=OutputInt;else if(p==0)visit=OutputChar;while(top>i)visit(*i++);puts("");return OK;}//StackTraverse}SqStack;Status OutputInt(SElemType e){printf("%d ",e);return OK;}Status OutputChar(SElemType e){printf("%c",e);return OK;}void CheckStackCode(){typedef int SElemType;int i;SqStack S;SElemType e;if(S.InitStack()==OK){for(i=1;i<=12;i++)S.Push(i);}printf("栈中元素依次为：");S.StackTraverse(1);S.Pop(e);printf("弹出的栈顶元素 e=%d\n",e);printf("栈空否：%d(1：空 0：否)\n",S.StackEmpty());S.GetTop(e);printf("栈顶元素 e=%d 栈的长度为%d\n",e,S.StackLength());S.Push(13);printf("栈中元素依次为：");S.StackTraverse(1);S.GetTop(e);printf("栈顶元素 e=%d 栈的长度为%d\n",e,S.StackLength());S.DestroyStack();printf("销毁栈后，S.top=%d  S.base=%d  S.stacksize=%d\n",S.top,S.base,S.stacksize);}void Conversion()//对于输入的任意一个非负十进制整数n，打印输出与其等值的八进制数{typedef int SElemType;int n;SqStack S;SElemType e;S.InitStack();//构造空栈scanf("%d",&n);while(n){S.Push(n%8);n/=8;}while(!S.StackEmpty()){S.Pop(e);printf("%d",e);}puts("");S.DestroyStack();}void BracketsMatching()//三种括号()[]{}，检验是否合法{char s[N];SqStack S;SElemType e;int i;S.InitStack();scanf("%s",s);for(i=0;i<strlen(s);i++){if(s[i]=='(' || s[i]=='{' || s[i]=='[')S.Push(s[i]);else if(s[i]==')' || s[i]=='}' || s[i]==']'){if(S.GetTop(e) && ((e=='(' && s[i]==')') || (e=='[' && s[i]==']') || (e=='{' && s[i]=='}')))S.Pop(e);else{puts("NO");S.DestroyStack();return;}}}if(S.StackEmpty())puts("YES");else puts("NO");S.DestroyStack();}void LineEdit()//从终端接收一行并传送至调用过程的数据区，#为退格符，@为退行符{int i;char ch;SqStack S;SElemType e;S.InitStack();while(ch!=EOF){for(ch=getchar();ch!=EOF && ch!='\n';ch=getchar()){if(ch=='#')S.Pop(e);else if(ch=='@')S.ClearStack();else S.Push(ch);}S.StackTraverse(0);S.ClearStack();}S.DestroyStack();}int main(){#ifndef ONLINE_JUDGEWfreopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;//init();//for(scanf("%d",&cass);cass;cass--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s))/*while(~scanf("%d%d",&n,&m)){}*/CheckStackCode();//测试Stack代码正确性Conversion();//数制转换BracketsMatching();//括号匹配的检验LineEdit();//行编辑程序return 0;}/*////*/

马踏棋盘问题（NxN）

暴力(顺序栈实现)

////by coolxxx//#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<stack>#include<queue>#include<set>#include<bitset>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>//#include<stdbool.h>#include<math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-10)#define J 10000#define mod 1000000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#pragma comment(linker,"/STACK:1024000000,1024000000")#define N 8using namespace std;typedef long long LL;double anss;LL aans;int cas,cass;LL n,m,lll,ans;int a[N][N];bool u[N][N];int dx[]={-2,-1,1,2,2,1,-1,-2};int dy[]={1,2,2,1,-1,-2,-2,-1};const int OK=1;const int ERROR=0;const int INFEASIBLE=-1;const int STACK_INIT_SIZE=10000;//存储空间初始分配量const int STACKINCREMENT=10000;//存储空间分配增量typedef int Status;typedef struct{int x,y,dir;}SElemType;Status OutputInt(SElemType e);Status OutputChar(SElemType e);typedef struct{SElemType *base;//栈构造之前和销毁之后，base的值为NULLSElemType *top;//栈顶指针int stacksize;//当前已分配的存储空间，以元素为单位Status InitStack()//构造一个空栈S{base=(SElemType *)malloc(STACK_INIT_SIZE*sizeof(SElemType));if(!base)exit(OVERFLOW);//存储分配失败top=base;stacksize=STACK_INIT_SIZE;return OK;}//InitStackStatus DestroyStack()//销毁栈S，S不再存在{free(base);base=NULL;top=NULL;stacksize=0;return OK;}//DestroyStackStatus ClearStack()//把S置为空栈{top=base;return OK;}//ClearStackStatus StackEmpty()//若S为空栈，则返回true，否则返回false{if(top==base)return true;return false;}//StackEmptyint StackLength()//返回S的元素个数，即栈的长度{return top-base;}//StackLengthStatus GetTop(SElemType &e)//若栈不空，则用e返回S的栈顶元素，并返回OK，否则返回ERROR{if(top==base)return ERROR;e=*(top-1);return OK;}//GetTopStatus Push(SElemType e)//插入元素e为新的栈顶元素{if(top-base>=stacksize)//栈满，追加存储空间{base=(SElemType *)realloc(base,(stacksize+STACKINCREMENT)*sizeof(SElemType));if(!base)exit(OVERFLOW);//存储分配失败top=base+stacksize;stacksize+=STACKINCREMENT;}(*top++)=e;return OK;}//PushStatus Pop(SElemType &e)//若栈不空，则删除S的栈顶元素，并用e返回其值，并返回OK，否则返回ERROR{if(top==base)return ERROR;e=(*--top);return OK;}//PopStatus StackTraverse(int p)//从栈底到栈顶依次对栈中每个元素调用函数visit().一旦visit()失败则操作失败{SElemType *i=base;Status (*visit)(SElemType);if(p==1)visit=OutputInt;else if(p==0)visit=OutputChar;while(top>i)visit(*i++);puts("");return OK;}//StackTraverse}SqStack;Status OutputInt(SElemType e){printf("%d ",e);return OK;}Status OutputChar(SElemType e){printf("%c",e);return OK;}SqStack S;void print(){int i,j;for(i=0;i<N;i++,puts(""))for(j=0;j<N;j++)printf("%d ",a[i][j]);puts("");}int main(){#ifndef ONLINE_JUDGEWfreopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;int x,y,z,xx,yy;//init();//for(scanf("%d",&cass);cass;cass--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s))while(~scanf("%d%d",&n,&m)){mem(a,0);mem(u,0);S.InitStack();SElemType e,to;e.x=n,e.y=m,e.dir=-1;u[n][m]=1;S.Push(e);loop : while(S.top-S.base<N*N && !S.StackEmpty()){SElemType & now=*(S.top-1);for(++now.dir;now.dir<8;now.dir++){xx=now.x+dx[now.dir],yy=now.y+dy[now.dir];if(xx>=0 && xx<N && yy>=0 && yy<N && !u[xx][yy]){u[xx][yy]=1;to.x=xx,to.y=yy,to.dir=-1;S.Push(to);goto loop;}}S.Pop(e);u[e.x][e.y]=0;}if(S.StackEmpty()){puts("No Answer\n");continue;}SElemType *id;for(id=S.base,cas=0;id!=S.top;id++){e=(*id);a[e.x][e.y]=++cas;}print();S.DestroyStack();}return 0;}

马踏棋盘贪心优化

// //by coolxxx//#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<stack>#include<queue>#include<set>#include<bitset>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>//#include<stdbool.h>#include<math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-10)#define J 10000#define mod 1000000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#pragma comment(linker,"/STACK:1024000000,1024000000")#define N 8using namespace std;typedef long long LL;double anss;LL aans;int cas,cass;LL n,m,lll,ans;int a[N][N];bool u[N][N];int dx[]={-2,-1,1,2,2,1,-1,-2};int dy[]={1,2,2,1,-1,-2,-2,-1};struct xxx{int c,num;};bool cmp(xxx aa,xxx bb){return aa.c<bb.c;}void print(){int i,j;for(i=0;i<N;i++,puts(""))for(j=0;j<N;j++)printf("%d ",a[i][j]);puts("");}void cal(int x,int y,xxx d[]){int i,j,xx,yy,x1,y1;for(i=0;i<8;i++){d[i].c=0;d[i].num=i;xx=x+dx[i],yy=y+dy[i];if(xx>=0 && xx<N && yy>=0 && yy<N && !u[xx][yy]){d[i].c++;for(j=0;j<8;j++){x1=xx+dx[j],y1=yy+dy[j];if(x1>=0 && x1<N && y1>=0 && y1<N && !u[x1][y1]){d[i].c++;}}}}}void dfs(int x,int y,int top){if(top==65){print();lll++;return;}int i,j,xx,yy;xxx d[8];cal(x,y,d);sort(d,d+8,cmp);for(i=0;i<8;i++){if(d[i].c==0)continue;j=d[i].num;xx=x+dx[j],yy=y+dy[j];if(xx>=0 && xx<N && yy>=0 && yy<N && !u[xx][yy]){u[xx][yy]=1;a[xx][yy]=top;dfs(xx,yy,top+1);if(lll==cas)return;u[xx][yy]=0;a[xx][yy]=0;}}}int main(){#ifndef ONLINE_JUDGEW//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;int x,y,z,xx,yy;//init();//for(scanf("%d",&cass);cass;cass--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s))puts("输入起点坐标和需要的解的数量");while(~scanf("%d%d",&n,&m)){scanf("%d",&cas);j=clock();mem(a,0);mem(u,0);lll=0;a[n][m]=1;u[n][m]=1;dfs(n,m,2);k=clock();printf("用时 %.3lf s\n",0.001*(k-j));puts("输入起点坐标和需要的解的数量");}return 0;}

为了写的快就随便优化没怎么细想，如果觉得我的算法复杂度还是太高可以转http://blog.csdn.net/bone_ace/article/details/41579957