栈与队列的三个例子
1.小括号匹配
#define FORMATSTR "%c "
typedef char ElemType;
#define BufSize 128
#include "Stack.h"
int CheckBracket(char *str, int len)
{
int i = 0;
SQSTACK S;
char ch;
InitSqstack(&S, len);
while(str[i])
{
ch = str[i];
if(ch == '(')
PushS(&S, ch);
else if(ch == ')')
if(!PopS(&S, &ch))
{
DestroySqstack(&S);
return 0;
}
++i;
}
if(IsSqstackEmpty(S))
;
else
i = 0;
DestroySqstack(&S);
return i;
}
int main(void)
{
//栈示例 - 括号匹配
char string[BufSize];
gets(string);
if(CheckBracket(string, strlen(string)))
puts("Match!");
else
puts("Mismatch!");
return 0;
}
2.迷宫求解
#define FORMATSTR
typedef struct
{
int x, y, v; //坐标与方向
}ElemType;
#include "Stack.h"
int Map[10][10] =
{
1,1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,0,1,
1,0,1,1,1,0,0,0,0,1,
1,0,0,0,1,0,0,0,0,1,
1,0,1,0,0,0,1,0,0,1,
1,0,1,1,1,0,1,1,0,1,
1,1,0,0,0,0,0,0,0,1,
1,1,1,1,1,1,1,1,1,1,
};
int tox[5] = {0, 0,1,0,-1}; //偏正方向
int toy[5] = {0,-1,0,1, 0}; //偏正方向
int main(void)
{
//链式栈示例 - 迷宫求解，给定的迷宫，输出所有解。
LINKSTACK S;
ElemType e;
int x, y, v; //当前坐标
int ox, oy; //修正坐标
int i, j, k = 0;
e.x = 1; e.y = 1; //出发点
e.v = 0; //方向
InitLinkstack(&S);
PushL(S, e);
Map[1][1] = 8;
while(!IsLinkstackEmpty(S))
{
PopL(S, &e);
x = e.x; y = e.y; v = e.v + 1;
//back from here
if(e.v > 0)
Map[y+toy[e.v]][x+tox[e.v]] = 0;
while(v <= 4)
{
ox = x + tox[v];
oy = y + toy[v];
if(ox == 8 && oy == 9) //终点
{
printf("Answer %d:\n", ++k);
for(i = 0; i < 10; ++i)
{
for(j = 0; j < 10; ++j)
printf("%d ", Map[i][j]);
putchar(10);
}
++v;
}
else if(ox > 0 && oy > 0 && !Map[oy][ox])
{
e.x = x; e.y = y; e.v = v;
PushL(S, e);
x = ox, y = oy; v = 1;
Map[y][x] = 8;
}
else
++v;
}
}
DestroyLinkstack(&S);
return 0;
}
3.最短路径
#define FORMATSTR
typedef struct
{
int x, y, pre; //坐标与方向
}ElemType;
#include "Queue.h"
int Map[10][10] =
{
1,1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,1,0,0,0,1,0,1,
1,0,0,0,0,1,1,0,0,1,
1,0,1,1,1,0,0,0,0,1,
1,0,0,0,1,0,0,0,0,1,
1,0,1,0,0,0,1,0,0,1,
1,0,1,1,1,0,1,1,0,1,
1,1,0,0,0,0,0,0,0,1,
1,1,1,1,1,1,1,1,1,1,
};
int tox[5] = {0, 0,1,0,-1}; //偏正方向
int toy[5] = {0,-1,0,1, 0}; //偏正方向
int main(void)
{
//队列示例 - 迷宫的最短路径
SQQUEUE Q;
ElemType e;
int x, y, v; //当前坐标
int ox, oy; //修正坐标
int i, j, k = 0, step = 2; //从02开始
e.x = 1; e.y = 1; //出发点
e.pre = 2; //前驱
InitSqqueue(&Q, 10*10);
EnSqqueue(&Q, e);
Map[1][1] = step;
while(!IsSqqueueEmpty(Q)) //BFS
{
DeSqqueue(&Q, &e);
x = e.x; y = e.y; v = e.pre += 1;
for(i = 1; i <= 4; ++i)
{
e.x = x + tox[i]; e.y = y + toy[i];
if(Map[e.y][e.x] == 0)
{
EnSqqueue(&Q, e);
Map[e.y][e.x] = e.pre;
if(e.x == 8 && e.y == 8) //终点
{
printf("The Shortest Path is:\n");
for(i = 0; i < 10; ++i)
{
for(j = 0; j < 10; ++j)
printf("%02d ", Map[i][j]);
putchar(10);
}
DestroySqqueue(&Q);
return 0; //找到就停
}
}
}
}
DestroySqqueue(&Q);
return 0;
}
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