简易双人五子棋

来源：互联网发布：oracle 实例端口编辑：程序博客网时间：2024/04/30 01:37
在最初学习了C语言和数据结构后就尝试着写写一个较为高端的程序，这个高端也是相对于当时掌握的只是来说的，于是我便写了这个简易小游戏，双人五子棋。由于水平有限，写的很粗糙，不过不管怎么说也是第一个独自完成的大点的程序,还有就是这个调试环境为DEV C++。下面是这个程序的代码：
#include <windows.h>#include <stdlib.h>#include <stdio.h>#include <string.h>#include <conio.h>#define STACK_SIZE 81#define TRUE 1#define FALSE 0typedef int ElemType;typedef struct{int qp[15][15];int sum[5];int ch;int yon; }data;typedef struct Node{ ElemType data[10]; int top;}seqstack; void initStack(seqstack *s) {s->top = -1;}int Empty(seqstack *s) {return (s->top == -1);}int Full(seqstack *s) {return (s->top == STACK_SIZE - 1);}int push(seqstack *s, ElemType elem) {if(Full(s))return FALSE;s->top++;s->data[s->top] = elem;return TRUE;}int pop(seqstack *s, ElemType *elem) {if(Empty(s))return FALSE;*elem = s->data[s->top];s->top--;return TRUE;}void wqizi(){printf("●");} void bqizi(){printf("○");} void gotoxy(int x,int y)  //通过坐标获取光标位置 {COORD pos;pos.X=x;pos.Y=y;SetConsoleCursorPosition(GetStdHandle(STD_OUTPUT_HANDLE),pos);}  void qipan(  )//棋盘界面 {   int MAX=29; int i,j;     for( i= 1; i<= MAX; ++i )    {        for( j= 1; j<=MAX; ++j )        {            if( i== 1 )            {                if( j== 1 )                    printf( "┏" );                  else if( j== MAX )                    printf( "┓\n" );                else if( j%2 )                    printf( "┳" ); // 横向占两个坐标位，竖向占一个坐标位                 else                    printf( "━" );            }            else if( i== MAX )            {                if( j== 1 )                    printf( "┗" );                else if( j== MAX )                    printf( "┛\n" );                else if( j%2 )                    printf( "┻" );                else                    printf( "━" );            }            else            {                if( j== 1 )                {                    if( i% 2 )                        printf( "┣" );                    else                        printf( "┃" );                }                else if( j== MAX )                {                    if( i% 2 )                        printf( "┫\n" );                    else                        printf( "┃\n" );                }                else                {                    if( i% 2 )                    {                        if( j% 2 )                            printf( "╋" );                        else                            printf( "━" );                    }                    else                    {                        if( j% 2 )                            printf( "┃" );                        else                              printf( "  " );                     }                }            }        }    }    gotoxy(60,12);printf("上下左右分别通");gotoxy(60,14);printf("过W,S,A,D按键");gotoxy(60,16);printf("来控制,落子通");gotoxy(60,18);printf("过按键Y来实现");}judgeL(seqstack *le,data *da,int x,int y,int xz)//判断水平方向是否满足获胜条件 {int r1,m;    r1=x;da->sum[1]=0;  initStack(le);while(da->qp[r1][y]==xz){  r1--; } m=r1+1;while(da->qp[m][y]==xz){  push(le, da->qp[m][y]);  m++;}while(!Empty(le)) {int tmp;pop(le, &tmp);da->sum[1]=da->sum[1]+tmp;}}judgeV(seqstack *le,data *da,int x,int y,int xz)//判断竖直方向是否满足获胜条件 {int r1,m;    r1=y;da->sum[2]=0;  initStack(le);while(da->qp[x][r1]==xz){  r1--; } m=r1+1;while(da->qp[x][m]==xz){  push(le, da->qp[x][m]);  m++;}while(!Empty(le)) {int tmp;pop(le, &tmp);da->sum[2]=da->sum[2]+tmp;}}judge45(seqstack *le,data *da,int x,int y,int xz)//判断向左45度是否满足获胜条件 {int r1,r2,m1,m2;    r1=x;    r2=y;da->sum[3]=0;  initStack(le);while(da->qp[r1][r2]==xz){  r1--;  r2--; } m1=r1+1; m2=r2+1;while(da->qp[m1][m2]==xz){  push(le, da->qp[m1][m2]);  m1++;  m2++;}while(!Empty(le)) {int tmp;pop(le, &tmp);da->sum[3]=da->sum[3]+tmp;}}judge135(seqstack *le,data *da,int x,int y,int xz)//判断向右45度是否满足获胜条件 {int r1,r2,m1,m2;    r1=x;    r2=y;da->sum[4]=0;  initStack(le);while(da->qp[r1][r2]==xz){  r1++;  r2--; } m1=r1-1; m2=r2+1;while(da->qp[m1][m2]==xz){  push(le,da->qp[m1][m2]);  m1--;  m2++;}while(!Empty(le)) {int tmp;pop(le, &tmp);da->sum[4]=da->sum[4]+tmp; }}void weizhi(data *da,int x,int y)//用来判断此位置是否已经落子 {if(da->qp[x][y]==1 || da->qp[x][y]==2){  gotoxy(65,10);  printf("此位置不合法");  Sleep(1000);  gotoxy(65,10);  printf("            ");      fflush(stdout);    da->ch=1; }}zlyj(seqstack *le,data *da)//让用户选择是否再来一局 {int s=0;da->yon=0;gotoxy(20,9);printf("按Y再来一局,按N退出");da->yon=getch();while(da->yon!='Y' && da->yon!='y' && da->yon!='N' && da->yon!='n'){da->yon=getch();}if(da->yon=='Y' || da->yon=='y'){gotoxy(0,0);qipan();dire(le,da);}    if(da->yon=='N' || da->yon=='n')da->yon=1;}void dire(seqstack *le,data *da)//用来控制棋子的移动及落子操作的实现 {int x=0,y=0,a,b,n=1,pd=2;int i,j,m,o[3]={0,0,0};gotoxy(65,5);printf("执黑子行");gotoxy(x,y);for(i=0;i<=15;i++)for(j=0;j<=15;j++)da->qp[i][j]=0;da->qp[x][y]=0;while(1){int sum1,sum2;a=getch();b=pow(-1,n);    if(a=='a' || a=='A')    {    if(x<1)continue;     gotoxy(4*(x-1),y*2);    x=x-1;    pd=1;    }    if(a=='d' || a=='D')    {    if(x>13)    continue;       gotoxy((x+1)*4,y*2);    x=x+1;    pd=1;    }    if(a=='w' || a=='W')    {    if(y<1)continue;    gotoxy(x*4,(y-1)*2);    y=y-1;    pd=1;    }    if(a=='s' || a=='S')    {    if(y>13)    continue;    gotoxy(x*4,(y+1)*2);    y=y+1;    pd=1;    }    if(b==-1 && (a=='y' || a=='Y') && pd==1)      { weizhi(da,x,y);if(da->ch==1){da->ch=0;continue;}      bqizi();      o[1]++;      n++;      pd=0;      da->qp[x][y]=1;      gotoxy(65,5);      printf("执白子行");gotoxy(x*4+2,y*2); judgeL(le,da,x,y,1);      judgeV(le,da,x,y,1);      judge45(le,da,x,y,1);      judge135(le,da,x,y,1);      if(da->sum[1]>5 || da->sum[2]>5 || da->sum[3]>5 || da->sum[4]>5)      {      gotoxy(20,5);      printf("由禁手规则,白棋获胜!");       zlyj(le,da);      if(da->yon==1)return 0;      }      if(da->sum[1]==5 || da->sum[2]==5 || da->sum[3]==5 || da->sum[4]==5)      {      gotoxy(25,5);printf("黑棋胜！");      zlyj(le,da);      if(da->yon==1)return 0;      }      }    if(b==1 && (a=='y' || a=='Y') && pd==1)      {weizhi(da,x,y);if(da->ch==1){da->ch=0;continue;}      wqizi();      o[0]++;      n++;      pd=0;      da->qp[x][y]=2;      gotoxy(65,5);      printf("执黑子行");      gotoxy(x*4+2,y*2);      judgeL(le,da,x,y,2);      judgeV(le,da,x,y,2);      judge45(le,da,x,y,2);      judge135(le,da,x,y,2);      if(da->sum[1]>=10 || da->sum[2]>=10 || da->sum[3]>=10 || da->sum[4]>=10)      {        gotoxy(25,7);  printf("白棋胜！");        zlyj(le,da);        if(da->yon==1)return 0;      }      }      if(pd==2 && (a=='y' || a=='Y'))      {      bqizi();      o[2]=1;      n++;      pd=0;      da->qp[x][y]=1;      gotoxy(65,5);      printf("执白子行");      gotoxy(x*4+2,y*2);      }           if(o[0]+o[1]+o[2]==225)      {      gotoxy(25,7);      printf("棋局已满");    zlyj(le,da);      if(da->yon==1)return 0;          }}}int main(){    gotoxy(32,10);    printf("欢迎使用双人对战五子棋系统!");    Sleep(2000);    system("cls");    fflush(stdout);seqstack le;data da;system("mode con cols=80 lines=40");    qipan();    dire(&le,&da);}
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