小白书上经典DFS题目。

1. 递归实现

// from: https://www.cnblogs.com/huaszjh/p/4686092.html

#include <stdio.h>
#include <string.h>
#define maxn 105
unsigned char data[maxn][maxn];
int m, n, vis[maxn][maxn]; void dfs(int x, int y, int ans) {
if (x < 0 || x >= m || y < 0 || y >= n) return; //出界
if (vis[x][y] > 0 || data[x][y] == '*') return; //非'@'或已经访问
vis[x][y] = ans; //连通分量编号
for (int k = -1; k <= 1; k++) {
for (int t = -1; t <= 1; t++) {
if (k != 0 || t != 0) { //自身格子不需要重复判断
dfs(x + k, y + t, ans);
}
}
}
} #define DEBUG
int main() {
#ifdef DEBUG
const char* input_txt_pth = "F:/zhangzhuo/debug/OJ/UVA-572.txt";
freopen(input_txt_pth, "r", stdin);
#endif int i, j;
while (scanf("%d %d", &m, &n) && m &&n) {
int count = 0; //连通块
memset(vis, 0, sizeof(vis));
for (i = 0; i < m; i++) {
scanf("%s", data[i]);
}
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
//对未访问且为`@`的格子进行访问
if (vis[i][j] == 0 && data[i][j] == '@') {
dfs(i, j, ++count);
}
}
}
printf("%d\n", count);
#ifdef DEBUG
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
printf("%3d", vis[i][j]);
}
printf("\n");
}
printf("\n");
#endif
}
return 0;
}

2. 递归dfs函数用迭代实现

每个节点的dfs递归调用,改成用stack容器就地计算,是个while循环,本质上还是栈,但是避免了递归时嵌套产生的开销造成的潜在风险。

C++的stack、vector容器用起来比较顺手。另外就是把坐标简单封装为一个结构体。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
#include <vector> typedef struct Coord {
char x, y;
} Coord; #define DEBUG
int main() {
#ifdef DEBUG
const char* input_txt_pth = "F:/zhangzhuo/debug/OJ/UVA-572.txt";
freopen(input_txt_pth, "r", stdin);
#endif int m, n, i, j; #define maxn 105
unsigned char data[maxn][maxn];
int vis[maxn][maxn]; while (scanf("%d %d", &m, &n) && m &&n) {
int count = 0; //连通块
memset(vis, 0, sizeof(vis));
for (i = 0; i < m; i++) {
scanf("%s", data[i]);
} std::stack<Coord> stk;
Coord cd;
std::vector<Coord>offset;
cd.x = -1; cd.y = -1; offset.push_back(cd);
cd.x = -1; cd.y = 0; offset.push_back(cd);
cd.x = -1; cd.y = 1; offset.push_back(cd);
cd.x = 0; cd.y = -1; offset.push_back(cd);
cd.x = 0; cd.y = 1; offset.push_back(cd);
cd.x = 1; cd.y = -1; offset.push_back(cd);
cd.x = 1; cd.y = 0; offset.push_back(cd);
cd.x = 1; cd.y = 1; offset.push_back(cd); for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
cd.x = i; cd.y = j;
if (vis[cd.x][cd.y] > 0 || data[cd.x][cd.y] != '@') continue;
count++; stk.push(cd);
while (!stk.empty()) {
cd = stk.top();
stk.pop();
vis[cd.x][cd.y] = count; Coord tmp;
for (size_t k = 0; k < offset.size(); k++) {
tmp.x = cd.x + offset[k].x;
tmp.y = cd.y + offset[k].y;
if (tmp.x < 0 || tmp.x >= m || tmp.y < 0 || tmp.y >= n) continue;
if (vis[tmp.x][tmp.y] > 0 || data[tmp.x][tmp.y] != '@') continue;
stk.push(tmp);
}
}
}
} printf("%d\n", count); #ifdef DEBUG
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
printf("%3d", vis[i][j]);
}
printf("\n");
}
printf("\n");
#endif
}
return 0;
}

3.纯C,DFS非递归,自定义栈ADT,函数指针

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string.h> typedef struct Coord Coord;
struct Coord {
char x, y;
}; typedef struct CoordOffset CoordOffset;
struct CoordOffset {
size_t num;
int* x;
int* y;
}; typedef struct ListNode ListNode;
struct ListNode
{
ListNode* next;
void* data;
}; typedef struct Stack Stack; struct Stack {
ListNode* head;
size_t len;
void(*push_coord)(Stack* stk, Coord* coord);
void (*pop_coord)(Stack* stk);
void (*top_coord)(Stack* stk, Coord* coord);
}; void stack_push_coord(Stack* stk, Coord* coord) {
ListNode* new_head = (ListNode*)malloc(sizeof(ListNode));
/* new_head->data = coord; */
new_head->data = (Coord*)malloc(sizeof(ListNode));
memcpy(new_head->data, coord, sizeof(Coord)); new_head->next = stk->head;
stk->head = new_head;
stk->len++;
} void stack_pop_coord(Stack* stk) {
if (stk->head != NULL) {
ListNode* new_head = stk->head->next;
free(stk->head->data);
free(stk->head);
stk->head = new_head;
stk->len--;
}
} void stack_top_coord(Stack* stk, Coord* coord) {
if (stk->head != NULL) {
Coord* t_coord = (Coord*)(stk->head->data);
coord->x = t_coord->x;
coord->y = t_coord->y;
}
} void make_stack(Stack** _stk) {
Stack* stk = (Stack*)malloc(sizeof(Stack));
stk->head = NULL;
stk->len = 0;
stk->push_coord = stack_push_coord;
stk->pop_coord = stack_pop_coord;
stk->top_coord = stack_top_coord; /* write back */
*_stk = stk;
} void free_stack(Stack* stk) {
ListNode* cur = stk->head;
ListNode* temp;
size_t i;
for (i = 0; i < stk->len; i++) {
temp = cur->next;
free(cur->data);
free(cur);
cur = temp;
}
free(stk);
stk = NULL;
} void make_8coord_offset(CoordOffset** _offset) {
CoordOffset* offset = (CoordOffset*)malloc(sizeof(CoordOffset));
offset->num = 8;
offset->x = (int*)malloc(sizeof(int)*offset->num);
offset->y = (int*)malloc(sizeof(int)*offset->num); offset->x[0] = -1; offset->y[0] = -1;
offset->x[1] = -1; offset->y[1] = 0;
offset->x[2] = -1; offset->y[2] = 1;
offset->x[3] = 0; offset->y[3] = -1;
offset->x[4] = 0; offset->y[4] = 1;
offset->x[5] = 1; offset->y[5] = -1;
offset->x[6] = 1; offset->y[6] = 0;
offset->x[7] = 1; offset->y[7] = 1; /* write back */
*_offset = offset;
} void free_coord_offset(CoordOffset* offset) {
if (offset) {
if (offset->x) {
free(offset->x);
offset->x = NULL;
}
if (offset->y) {
free(offset->y);
offset->y = NULL;
}
free(offset);
offset = NULL;
}
} /* #define DEBUG */
int main() {
#ifdef DEBUG
const char* input_txt_pth = "F:/zhangzhuo/debug/OJ/UVA-572.txt";
freopen(input_txt_pth, "r", stdin);
#endif int m, n, i, j;
size_t k; #define maxn 105
unsigned char data[maxn][maxn];
int vis[maxn][maxn]; /* here we use 8 neighbours */
CoordOffset* offset = NULL;
make_8coord_offset(&offset); while (scanf("%d %d", &m, &n) && m &&n) {
int count = 0; /* 连通块 */
memset(vis, 0, sizeof(vis));
for (i = 0; i < m; i++) {
scanf("%s", data[i]);
} /* std::stack<Coord> stk; */
Stack* stk;
make_stack(&stk); Coord cd; for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
cd.x = i; cd.y = j;
if (vis[cd.x][cd.y] > 0 || data[cd.x][cd.y] != '@') continue;
count++; /* stk.push(cd); */
stack_push_coord(stk, &cd);
/* while (!stk.empty()) { */
while(stk->len!=0) {
/* cd = stk.top(); */
/* stack_top_coord(stk, &cd); */
stk->top_coord(stk, &cd);
/* stk.pop(); */
/* stack_pop_coord(stk); */
stk->pop_coord(stk); vis[cd.x][cd.y] = count; Coord tmp;
for (k = 0; k < offset->num; k++) {
tmp.x = cd.x + offset->x[k];
tmp.y = cd.y + offset->y[k];
if (tmp.x < 0 || tmp.x >= m || tmp.y < 0 || tmp.y >= n) continue;
if (vis[tmp.x][tmp.y] > 0 || data[tmp.x][tmp.y] != '@') continue;
/* stk.push(tmp); */
/* stack_push_coord(stk, &tmp); */
stk->push_coord(stk, &tmp);
}
}
}
}
free_stack(stk); printf("%d\n", count); #ifdef DEBUG
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
printf("%3d", vis[i][j]);
}
printf("\n");
}
printf("\n");
#endif
} free_coord_offset(offset);
return 0;
}

4.DFS+并查集实现

#include <stdio.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h> int fa[10500];
int m, n, cnt, vis[105][105];
char mp[105][105];
int find(int x) {
if (fa[x] == x) return x;
fa[x] = find(fa[x]);
return fa[x];
} void merge(int x, int y) {
int fx = find(x);
int fy = find(y);
if (fx == fy) return;
fa[fx] = fy;
} void dfs(int x, int y, int fx, int fy) {
if (x < 0 || x >= m || y < 0 || y >= n) return;
if (vis[x][y] || mp[x][y] == '*') return;
vis[x][y] = 1;
/* cout<<"x || y || fx || fy : "<<x<<" || "<<y<<" || "<<fx<<" || "<<fy<<endl; */
if (fx != -1) {
merge(x*m + y, fx*m + fy);
}
int i, j;
for (i = -1; i < 2; i++) {
for (j = -1; j < 2; j++) {
if (!i && !j) continue;
dfs(x + i, y + j, x, y);
}
}
} /* #define LOCAL */
int main() {
#ifdef LOCAL
const char* input_txt = "F:/zhangzhuo/debug/OJ/UVA-572.txt";
freopen(input_txt, "r", stdin);
#endif
int i, j;
while (scanf("%d%d", &m, &n) == 2 && m && n) {
cnt = 0;
memset(vis, 0, sizeof(vis));
for (i = 0; i < m; i++) {
scanf("%s", mp[i]);
}
for (i = 0; i < 10500; i++) {
fa[i] = i;
}
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
if (!vis[i][j] && mp[i][j] == '@') {
dfs(i, j, -1, -1);
cnt++;
}
}
}
printf("%d\n", cnt); #ifdef LOCAL
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
printf("%3d", vis[i][j]);
}
printf("\n");
}
printf("\n");
#endif }
return 0;
}

5.DFS+并查集+不使用全局变量+简单封装为结构体

修改自 UVA572 (并查集解法) 。这种写法有点问题:已经用了dfs,dfs里用并查集多此一举,如果用并查集就不应该递归调用dfs。

#include <stdio.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h> typedef struct FSU_Node {
int p; /* parent id */
int rank;
int vis; /* group(connected component) id */
} FSU_Node; /*
get node's root id
@param x: node id
@param nodes: all nodes in map
*/
int fus_find(int x, FSU_Node* nodes) {
if (nodes[x].p == x) return x; nodes[x].p = fus_find(nodes[x].p, nodes);
return nodes[x].p;
} /*
merge two node groups
@param a: a node from one node group
@param b: a node from another node group
*/
void fus_union(int a, int b, FSU_Node* nodes)
{
int ra = fus_find(a, nodes); /* ra: root id of a */
int rb = fus_find(b, nodes); /* rb: root id of b */
if (ra == rb) {
return;
} if (nodes[ra].rank > nodes[rb].rank)
{
nodes[rb].p = ra;
}
else {
if (nodes[ra].rank == nodes[rb].rank)
{
nodes[rb].rank++;
}
nodes[ra].p = rb;
}
} typedef struct ImageSize {
int w, h;
} ImageSize; typedef struct Coord {
int row, col;
} Coord; void fus_dfs(const Coord* pt, const Coord* f_pt, FSU_Node* nodes, ImageSize* sz, unsigned char* mp) {
int row = pt->row;
int col = pt->col; int f_row = f_pt->row;
int f_col = f_pt->col; if (row < 0 || row >= sz->h || col < 0 || col >= sz->w) return; int id = row * sz->w + col;
int fid = f_row * sz->w + f_col; /* if (vis[id] || mp[id] == '*') return; */
if (nodes[id].vis || mp[id] == '*') return;
/* vis[id] = 1; */
nodes[id].vis = 1; if (f_row != -1) {
fus_union(id, fid, nodes);
} int i, j;
Coord neighbor;
for (i = -1; i < 2; i++) {
for (j = -1; j < 2; j++) {
if (!i && !j) continue;
neighbor.row = row + i;
neighbor.col = col + j;
fus_dfs(&neighbor, pt, nodes, sz, mp);
}
}
} /*#define LOCAL*/
int main() {
#ifdef LOCAL
const char* input_txt = "F:/zhangzhuo/debug/OJ/UVA-572.txt";
freopen(input_txt, "r", stdin);
#endif #define MAXN 105
int m, n, cnt, i, j; /* int vis[MAXN*MAXN]; */
unsigned char mp[MAXN*MAXN];
FSU_Node nodes[MAXN*MAXN]; int idx; while (scanf("%d%d", &m, &n) == 2 && m && n) {
cnt = 0;
/* memset(vis, 0, sizeof(int)*MAXN*MAXN); */
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
idx = i * n + j;
scanf(" %c", &mp[idx]);
/* printf("! %c !", mp[idx]); */
}
}
for (i = 0; i < m*n; i++) {
nodes[i].p = idx;
nodes[i].rank = 1;
nodes[i].vis = 0;
} ImageSize im_sz;
im_sz.h = m;
im_sz.w = n;
Coord pt;
Coord f_pt;
f_pt.row = -1;
f_pt.col = -1; for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
idx = i * n + j;
/* if (!vis[idx] && mp[idx] == '@') { */
if (!nodes[idx].vis && mp[idx] == '@') {
/* dfs(i, j, -1, -1); */
pt.row = i;
pt.col = j;
/* fus_dfs(&pt, &f_pt, nodes, &im_sz, vis, mp); */
fus_dfs(&pt, &f_pt, nodes, &im_sz, mp);
cnt++;
}
}
}
printf("%d\n", cnt); #ifdef LOCAL
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
idx = i * m + j;
/* printf("%3d", vis[idx]); */
printf("%c", mp[idx]);
}
printf("\n");
}
printf("\n");
#endif
} return 0;
}

这里的教训是,如果在双重for循环中使用变量x、y来表示坐标,容易把2维度坐标->1维坐标的计算算错。使用row,col能减少犯错可能;

另外就是数据读取,这里改成%c,则需要过滤掉换行符\n,方法是scanf时的格式串首部添加空格:scanf(" %c", &xx)

6. 并查集,去掉了DFS

思路:遍历每个像素点,每个像素点用并查集算法合并周边8邻域中为'@'的像素点。再次遍历,统计每个'@'像素对应的等价类(root节点)的值。第三次遍历,把第二次统计的值当中cnt数大于0的累计,就是区域个数。在统计连通域个数的时候顺带把每个连通域id(像素的parent值)修改为从1开始严格单调增的序列,开启LOCAL和LOCAL_DEBUG宏可以看到。

和通常用的模板写法略有差别,比如返回root的递归终止条件,比如root初值。

不得不说,uDebug是个好东西。

#include <stdio.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h> typedef struct FSU_Node {
int p; /* parent id */
int rank;
} FSU_Node; /*
get node's root id
@param x: node id
@param nodes: all nodes in map
*/ int fus_find(int x, FSU_Node* nodes) {
if (nodes[x].p == x) {
return x;
} nodes[x].p = fus_find(nodes[x].p, nodes);
return nodes[x].p;
} /*
merge two node groups
@param a: a node from one node group
@param b: a node from another node group
*/
void fus_union(int a, int b, FSU_Node* nodes)
{
int ra = fus_find(a, nodes); /* ra: root id of a */
int rb = fus_find(b, nodes); /* rb: root id of b */
if (ra == rb) {
return;
} if (nodes[ra].rank > nodes[rb].rank) {
nodes[rb].p = ra;
}
else {
if (nodes[ra].rank == nodes[rb].rank) {
nodes[rb].rank++;
}
nodes[ra].p = rb;
}
} /* #define LOCAL */
/* #define LOCAL_DEBUG */
int main() {
#ifdef LOCAL
const char* input_txt = "F:/zhangzhuo/debug/OJ/UVA-572.txt";
freopen(input_txt, "r", stdin);
#endif #define MAXN 105
int m, n, cnt, i, j, k; int shift_x[8] = { -1, -1, -1, 0, 0, 1, 1, 1 };
int shift_y[8] = { -1, 0, 1, -1, 1, -1, 0, 1 }; unsigned char mp[MAXN*MAXN];
FSU_Node nodes[MAXN*MAXN]; int idx; while (scanf("%d%d", &m, &n) == 2 && m && n) {
cnt = 0;
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
idx = i * n + j;
scanf(" %c", &mp[idx]);
}
}
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
idx = i * n + j;
nodes[idx].p = idx;
nodes[idx].rank = 1;
}
} for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
idx = i * n + j;
if (mp[idx] != '@') continue;
for (k = 0; k < 8; k++) {
int row = i + shift_x[k];
int col = j + shift_y[k];
int neighbor_idx = row * n + col;
if (row < 0 || row >= m || col < 0 || col >= n || mp[neighbor_idx] != '@') continue;
fus_union(idx, neighbor_idx, nodes);
}
}
} int bowl[MAXN*MAXN] = { 0 };
int label_cnt = 0;
for (i = 0; i < m*n; i++) {
if (mp[i] != '@') continue;
int t = fus_find(i, nodes);
nodes[i].p = t;
if (bowl[t] == 0) {
label_cnt++;
bowl[t] = label_cnt;
}
}
printf("%d\n", label_cnt); #ifdef LOCAL_DEBUG
/* print out debug info */
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
idx = i * n + j ;
if (mp[idx] == '@') {
/* printf("%3d", fus_find(idx, nodes)); */
/* printf("%3d", nodes[idx].p); */
printf("%3d", bowl[nodes[idx].p]);
}
else {
printf("%3c", '*');
}
}
printf("\n");
}
printf("\n");
#endif } return 0;
}

UVA572 Oil Deposits DFS求解的更多相关文章

  1. UVa572 Oil Deposits DFS求连通块

      技巧:遍历8个方向 ; dr <= ; dr++) ; dc <= ; dc++) || dc != ) dfs(r+dr, c+dc, id); 我的解法: #include< ...

  2. HDOJ(HDU).1241 Oil Deposits(DFS)

    HDOJ(HDU).1241 Oil Deposits(DFS) [从零开始DFS(5)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架 ...

  3. Oil Deposits(dfs)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  4. HDU 1241 Oil Deposits DFS(深度优先搜索) 和 BFS(广度优先搜索)

    Oil Deposits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...

  5. HDU 1241 Oil Deposits (DFS/BFS)

    Oil Deposits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tota ...

  6. HDU-1241 Oil Deposits (DFS)

    Oil Deposits Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total ...

  7. HDU_1241 Oil Deposits(DFS深搜)

    Problem Description The GeoSurvComp geologic survey company is responsible for detecting underground ...

  8. UVa 572 Oil Deposits(DFS)

     Oil Deposits  The GeoSurvComp geologic survey company is responsible for detecting underground oil ...

  9. [POJ] 1562 Oil Deposits (DFS)

    Oil Deposits Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16655   Accepted: 8917 Des ...

随机推荐

  1. leetcode No.500 键盘行 keyboard-row (Python3实现)

    题目描述 给定一个单词列表,只返回可以使用在键盘同一行的字母打印出来的单词.键盘如下图所示. 示例: 输入: ["Hello", "Alaska", " ...

  2. iOS - 判断程序每天只执行一次

    当进行操作的时候记录操作时间存在偏好设置当中,当再次点击的时候获取现在的时间然后和之前记录的时间进行比较.如果是一天那么就提示“今天已经操作过了”,如果不是一天,那么可以正常操作,然后记录操作时间.如 ...

  3. photoshop7.0 排版一寸照片、2寸照片

    说明:必须先照一张一寸电子照片,否则是无法做成 1.本例同样采用photoshop CS5制作,其它版本通用,这里采用上一教程“PS照片处理教程-制作一寸照片并排版”的处理效果图进行排版,首先在PS中 ...

  4. 深入Nginx之《常用参数配置技巧》

    常见参配置实战技巧 下面会讲解实战中应该怎么配置更为合理. 1.user 默认是nobody,如果使用nobody,Nginx在运行过程中会出现很多操作没有权限,比如写硬盘.一般都是用低于root级别 ...

  5. spring boot 指定启动端口

    spring boot 默认端口为8080 1.修改为指定端口 (1)修改配置文件 src/main/resources/application.properties server.port= (2) ...

  6. net core 环境部署的坑

    1.supervisor “no such file” error. 检查指令是否正确,路径.dotnet环境是否正常 2.Couldn‘t find a valid ICU package inst ...

  7. activiti学习2:示例工程activiti-explorer.war的使用

    目录 activiti学习2:示例工程activiti-explorer.war的使用 一.搭建开发环境 二.运行示例工程 三.示例工程功能演示 1. 创建流程图 2. 部署流程图 3. 启动流程 4 ...

  8. Java设计模式--观察者模式到监听器

    观察者模式是对象的行为模式.又叫做发布-订阅模式.模型-视图模式.源-监听器模式. 抽象主题角色:主题角色将所有对观察者对象的引用到保存在一个集合里,每个主题都可以拥有任意数量的观察者.抽象主题提供一 ...

  9. Ubuntu18.04LTS python3.6 cuda10.0 下安装低版本的pytorch

    Ubuntu18.04LTS python3.6 cuda10.0 下安装低版本的pytorch,运行Hypergraph Neural Networks(HGNN) https://github.c ...

  10. 字符串A转换到字符串B,只能一次一次转换,每次转换只能把字符串A中的一个字符全部转换成另一个字符,是否能够转换成功

    public class DemoTest { public static void main(String[] args) { System.)); } /** * 有一个字符串A 有一个字符串B ...