From da0755858e858e1234cfc57d2d1a5faed04bd199 Mon Sep 17 00:00:00 2001 From: Luthics Date: Sat, 2 Dec 2023 14:12:30 +0800 Subject: [PATCH] =?UTF-8?q?=E8=A1=A5=E5=85=85=E7=AC=AC=E5=8D=81=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 10/data.in | 8 +++ 10/main.c | 163 +++++++++++++++++++++++++++++++++++++++++++++++++++++ 13/bfs.c | 72 +++++++++++++++++++++++ 13/data.in | 8 +++ 13/dfs.c | 63 +++++++++++++++++++++ 13/main.c | 1 - 6 files changed, 314 insertions(+), 1 deletion(-) create mode 100644 10/data.in create mode 100644 10/main.c create mode 100644 13/bfs.c create mode 100644 13/data.in create mode 100644 13/dfs.c delete mode 100644 13/main.c diff --git a/10/data.in b/10/data.in new file mode 100644 index 0000000..f7b5941 --- /dev/null +++ b/10/data.in @@ -0,0 +1,8 @@ +4 +1 2 +5 2 +3 4 +666 9 +2 +1 3 +2 4 \ No newline at end of file diff --git a/10/main.c b/10/main.c new file mode 100644 index 0000000..94a5c76 --- /dev/null +++ b/10/main.c @@ -0,0 +1,163 @@ +// 1.问题描述 +// 设计一个一元稀疏多项式简单计算器。 +// 2.基本要求 +// 一元稀疏多项式简单计算器的基本功能是： +// (1)输入并建立多项式； +// (2)输出多项式，输出形式为整数序列： +// n,c1,e1,c2,e2,…,cn,en, +// 其中n是多项式的项数，ci,ei分别是第i项的系数和指数，序列按指数降序排列； +// (3)多项式a和b相加，建立多项式a+b； +// (4)多项式a和b相减，建立多项式a-b； +// (5)计算多项式在x处的值；即给定x值，计算多项式值。 +// 3.实现提示 +// 用带表头结点的单链表存储多项式，多项式的项数存放在头结点中。 + +#include +#include +#include + +typedef struct Node { + int c, e; + struct Node *next; +} Node; + +typedef struct List { + Node head; + int len; +} List; + +List *init_list() { + List *l = (List *)malloc(sizeof(List)); + l->head.next = NULL; + l->len = 0; + return l; +} + +// 降序插入 +void insert(List *l, int c, int e) { + Node *p = &(l->head); + while (p->next && p->next->e > e) { + p = p->next; + } + if (p->next && p->next->e == e) { + p->next->c += c; + if (p->next->c == 0) { + Node *q = p->next; + p->next = q->next; + free(q); + l->len--; + } + } else { + Node *q = (Node *)malloc(sizeof(Node)); + q->c = c, q->e = e; + q->next = p->next; + p->next = q; + l->len++; + } +} + +// 只输出 +void output(List *l) { + Node *p = l->head.next; + printf("%d,", l->len); + while (p) { + printf("%d,%d", p->c, p->e); + p = p->next; + if (p) + printf(","); + } + printf("\n"); +} + +// 只输出 +void plus(List *l1, List *l2) { + Node *p1 = l1->head.next, *p2 = l2->head.next; + while (p1 && p2) { + if (p1->e > p2->e) { + printf("%d,%d,", p1->c, p1->e); + p1 = p1->next; + } else if (p1->e < p2->e) { + printf("%d,%d,", p2->c, p2->e); + p2 = p2->next; + } else { + printf("%d,%d,", p1->c + p2->c, p1->e); + p1 = p1->next; + p2 = p2->next; + } + } + while (p1) { + printf("%d,%d,", p1->c, p1->e); + p1 = p1->next; + } + while (p2) { + printf("%d,%d,", p2->c, p2->e); + p2 = p2->next; + } + printf("\n"); +} + +// 只输出 +void minus(List *l1, List *l2) { + Node *p1 = l1->head.next, *p2 = l2->head.next; + while (p1 && p2) { + if (p1->e > p2->e) { + printf("%d,%d,", p1->c, p1->e); + p1 = p1->next; + } else if (p1->e < p2->e) { + printf("%d,%d,", -p2->c, p2->e); + p2 = p2->next; + } else { + printf("%d,%d,", p1->c - p2->c, p1->e); + p1 = p1->next; + p2 = p2->next; + } + } + while (p1) { + printf("%d,%d,", p1->c, p1->e); + p1 = p1->next; + } + while (p2) { + printf("%d,%d,", -p2->c, p2->e); + p2 = p2->next; + } + printf("\n"); +} + +int calc(List *l, int x) { + int sum = 0; + Node *p = l->head.next; + while (p) { + sum += p->c * pow(x, p->e); + p = p->next; + } + return sum; +} + +int main() { + // freopen("data.in", "r", stdin); + List *l = init_list(); + List *l2 = init_list(); + int n1, n2, c, e; + scanf("%d", &n1); + for (int i = 0; i < n1; i++) { + scanf("%d%d", &c, &e); + insert(l, c, e); + } + scanf("%d", &n2); + for (int i = 0; i < n2; i++) { + scanf("%d%d", &c, &e); + insert(l2, c, e); + } + output(l); + output(l2); + plus(l, l2); + minus(l, l2); + while (1) { + printf("请输入x的值："); + scanf("%d", &c); + if (c == -1) + break; + printf("%d\n", calc(l, c)); + } + return 0; +} diff --git a/13/bfs.c b/13/bfs.c new file mode 100644 index 0000000..4c3dd35 --- /dev/null +++ b/13/bfs.c @@ -0,0 +1,72 @@ +// // Ref: P1605 迷宫 +// 1、问题描述： +// 迷宫实验是取自心理学的一个古典实验。在该实验中，把一只老鼠从一个无顶大盒子的门放入，在盒中设置了许多墙，对行进方向形成了多处阻挡。盒子仅有一个出口，在出口处放置一块奶酪，吸引老鼠在迷宫中寻找道路以到达出口。对同一只老鼠重复进行上述实验，一直到老鼠从入口到出口，而不走错一步。老鼠经多次试验终于得到它学习走迷宫的路线。 +// 2、设计功能要求： +// 迷宫由m行n列的二维数组设置，0表示无障碍，1表示有障碍。设入口为（1，1），出口为（m，n），每次只能从一个无障碍单元移到周围四个方向上任一无障碍单元。编程实现对任意设定的迷宫，求出一条从入口到出口的通路，或得出没有通路的结论。 +// 算法输入：代表迷宫入口的坐标 +// 算法输出：穿过迷宫的结果。 算法要点：创建迷宫，试探法查找路。 +#include +#include +#include + +int m, n; +int maze[105][105]; +int vis[105][105]; + +typedef struct { + int x, y; +} Node; + +Node *q[10005]; +int head, tail; + +int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; + +void push(Node *node) { q[tail++] = node; } + +Node *front() { return q[head]; } + +void pop() { head++; } + +int empty() { return head == tail; } + +int main() { + freopen("data.in", "r", stdin); + scanf("%d%d", &m, &n); + for (int i = 1; i <= m; i++) { + for (int j = 1; j <= n; j++) { + scanf("%d", &maze[i][j]); + } + } + Node *start = (Node *)malloc(sizeof(Node)); + start->x = 1; + start->y = 1; + push(start); + vis[1][1] = 1; + while (!empty()) { + Node *cur = front(); + pop(); + for (int i = 0; i < 4; i++) { + int nx = cur->x + dir[i][0]; + int ny = cur->y + dir[i][1]; + if (nx < 1 || nx > m || ny < 1 || ny > n) { + continue; + } + if (maze[nx][ny] == 0 && !vis[nx][ny]) { + Node *next = (Node *)malloc(sizeof(Node)); + next->x = nx; + next->y = ny; + push(next); + vis[nx][ny] = 1; + } + } + } + + if (vis[m][n]) { + printf("Yes\n"); + } else { + printf("No\n"); + } + + return 0; +} \ No newline at end of file diff --git a/13/data.in b/13/data.in new file mode 100644 index 0000000..4864c00 --- /dev/null +++ b/13/data.in @@ -0,0 +1,8 @@ +7 6 +0 0 1 1 0 0 +1 0 0 0 0 0 +1 0 0 0 0 1 +1 0 0 0 0 1 +1 1 0 0 0 0 +1 1 1 1 0 0 + diff --git a/13/dfs.c b/13/dfs.c new file mode 100644 index 0000000..cc07b96 --- /dev/null +++ b/13/dfs.c @@ -0,0 +1,63 @@ +#include +#include +#include + +int m, n; +int maze[105][105]; + +// 存储当前路径 +char *path[10005]; +int path_len = 0; + +void dfs(int x, int y) { + if (x == m && y == n) { + for (int i = 0; i < path_len; i++) { + printf("%s", path[i]); + if (i != path_len - 1) { + printf("->"); + } + } + printf("\n"); + return; + } + maze[x][y] = 1; + if (x + 1 <= m && maze[x + 1][y] == 0) { + path[path_len] = (char *)malloc(sizeof(char) * 105); + sprintf(path[path_len++], "(%d, %d)", x + 1, y); + dfs(x + 1, y); + path_len--; + } + if (y + 1 <= n && maze[x][y + 1] == 0) { + path[path_len] = (char *)malloc(sizeof(char) * 105); + sprintf(path[path_len++], "(%d, %d)", x, y + 1); + dfs(x, y + 1); + path_len--; + } + if (x - 1 >= 1 && maze[x - 1][y] == 0) { + path[path_len] = (char *)malloc(sizeof(char) * 105); + sprintf(path[path_len++], "(%d, %d)", x - 1, y); + dfs(x - 1, y); + path_len--; + } + if (y - 1 >= 1 && maze[x][y - 1] == 0) { + path[path_len] = (char *)malloc(sizeof(char) * 105); + sprintf(path[path_len++], "(%d, %d)", x, y - 1); + dfs(x, y - 1); + path_len--; + } +} + +int main() { + freopen("data.in", "r", stdin); + scanf("%d%d", &m, &n); + for (int i = 1; i <= m; i++) { + for (int j = 1; j <= n; j++) { + scanf("%d", &maze[i][j]); + } + } + path[path_len] = (char *)malloc(sizeof(char) * 105); + strcpy(path[path_len++], "(1, 1)"); + dfs(1, 1); + + return 0; +} \ No newline at end of file diff --git a/13/main.c b/13/main.c deleted file mode 100644 index 60ac95e..0000000 --- a/13/main.c +++ /dev/null @@ -1 +0,0 @@ -// Ref: P1605 迷宫 \ No newline at end of file