POJ 1979 Red and Black(C语言堆栈实现)

一、题目信息

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 17456 Accepted: 9198

Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles. 

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20. 

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows. 

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 
The end of the input is indicated by a line consisting of two zeros. 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

Sample Output

4559613

二、算法分析

 本题要求包含@的连通区域，跟图形学中种子填充算法类似，采用堆栈，将@所在位置作为种子，依次填充该种子前后左右，直到堆栈为空。

三、参考代码

（采用GCC提交的话，不要包含<malloc.h><memory.h>等头文件）

#include <stdio.h>#include <string.h>//#include <malloc.h>struct dirNode{    unsigned x;    unsigned y;    struct dirNode * next;};struct stack{    struct dirNode *top;};struct stack * Stack_init(){    struct stack *p;    p = (struct stack *)malloc(sizeof(struct stack));    if(p != NULL)        p->top = NULL;    return p;}struct stack * Stack_insert(struct stack *s,int x,int y){    struct dirNode *p;    p = (struct dirNode *)malloc(sizeof(struct dirNode));    if(p != NULL)    {        p->x = x;        p->y = y;        p->next = NULL;    }    p->next = s->top;    s->top = p;    return s;}struct stack * Stack_pop(struct stack *s){    struct dirNode *p;    if (s == NULL || s->top == NULL)        return s;    p = s->top;    s->top = p->next;    free(p);    return s;}int main(){    int W,H,i,j,room[20][20],res = 0,ii,jj,tmpi,tmpj;    short cal[4][2] = {-1,0,0,+1,+1,0,0,-1};    char tmp;    struct dirNode *stack = NULL;    while(scanf("%d%d",&W,&H) && W && H)    {        res = 0;        struct stack *s = Stack_init();        getchar();        for( i = 0 ; i < H ;i++)        {            for( j = 0 ; j < W; )            {                tmp = getchar();                room[i][j] = (tmp == '#') ? 0 : 1;                            if (tmp == '@')                {                    s = Stack_insert(s,i,j);                    room[i][j] = 0;                    res += 1;                }                j++;            }            getchar();        }        while(s->top!= NULL)        {            ii = s->top->x;            jj = s->top->y;            s = Stack_pop(s);            for(i = 0;i < 4 ;i++)            {                tmpi = ii + cal[i][0];                tmpj = jj + cal[i][1];                if( (tmpi < H)&&(tmpi >=0) &&(tmpj < W) && (tmpj >=0 ))                {                    if(room[tmpi][tmpj] == 1)                    {                        room[tmpi][tmpj] = 0;                        s = Stack_insert(s,tmpi,tmpj);                        res++;                    }                }            }        }        printf("%d\n",res);    }      //system("pause");    return 0;}