函数+编程题

7-1 交换二叉树中每个结点的左孩子和右孩子

#include<stdio.h>
#include<stdlib.h>
typedef struct Node {
    char data;
    struct Node *left, *right;
} Node, *Tree;

Tree create() {
    char ch;
    Tree tree = NULL;
    if (scanf("%c", &ch)) {//当有输入时
        if (ch != '#') {
            tree = (Tree)malloc(sizeof(Node));
            tree->data = ch;
            tree->left = create();
            tree->right = create();
        } else {
            tree = NULL;
        }
    }
    return tree;
}

void order(Tree tree) {
    if (!tree)
        return;
    order(tree->left);
    printf("%c", tree->data);
    order(tree->right);
}

void swap(Tree tree) {
    if (!tree)
        return;
    if (!tree->left && !tree->right)
        return;
    Tree tmp = tree->left;
    tree->left = tree->right;
    tree->right = tmp;
    swap(tree->left);
    swap(tree->right);
}

int main() {
    Tree tree = create();
    order(tree);
    printf("\n");
    swap(tree);
    order(tree);
    return 0;
}

7-2 两个有序链表序列的合并

一开始用的是数组方法

#include<stdio.h>
#include<stdlib.h>
int s1[1010], s2[1010];
int x, cnta, cntb;
int cmp(const void *a, const void *b){
    return(*(int*)a - *(int*)b);
}
int main(){
    while(scanf("%d", &x) != EOF){
        if(x == -1) break;
        else s1[++ cnta] = x;
    }
    while(!scanf("%d", &x) != EOF){
        if(x == -1) break;
        else s2[++ cntb] = x;
    }
    for(int i = cnta + 1; i <= cnta + cntb; i ++ ){
        s1[i] = s2[i - cnta];
    }
    qsort(s1, cnta + cntb, sizeof(s1[0]), cmp);
    if(cnta + cntb == 0){
        printf("NULL");
        return 0;
    }
    for(int i = 1; i <= cnta + cntb; i ++ ){
        if(i != 1) printf(" ");
        printf("%d", s1[i]);
    }

后来发现在大规模输入的时候不能过

于是就借鉴网上的ac了嘤嘤嘤嘤QAQ

#include<stdio.h>
#include<stdlib.h>
typedef struct Node *List;
struct Node{
    int data;
    struct Node *Next;
};
List InitList(){
    List l;
    l = (List)malloc(sizeof(struct Node));
    if(!l) return NULL;
    l->Next = NULL;
    return l;
}
void read(List l){
    List t;
    int data;
    scanf("%d", &data);
    while(data > 0){
        t = (List)malloc(sizeof(struct Node));
        if(!t) return ;
        t->data = data;
        t->Next = NULL;
        l->Next = t;
        l = l->Next;
        scanf("%d", &data);
    }
    return ;
}
void combine(List l1, List l2, List l3){
    l1 = l1->Next;
    l2 = l2->Next;
    while(l1 != NULL && l2 != NULL){
        if(l1->data > l2->data){
            l3->Next = l2;
            l2 = l2->Next;
        }
        else{
            l3->Next = l1;
            l1 = l1->Next;
        }
        l3 = l3->Next;
    }
    if(l1 == NULL && l2 == NULL) return ;
    if(l1 != NULL) l3->Next = l1;
    else l3->Next = l2;
    return ;
}
void print(List l){
    l = l->Next;
    if(l == NULL){
        printf("NULL");
        return ;
    }
    while(l){
        if(l->Next == NULL) printf("%d", l->data);
        else printf("%d ", l->data);
        l = l->Next;
    }
}
int main(){
    List l1, l2, l3;
    l1 = InitList();
    l2 = InitList();
    l3 = InitList();
    read(l1);
    read(l2);
    combine(l1, l2, l3);
    print(l3);
    return 0;
}

6-4 链表逆置

struct ListNode *reverse( struct ListNode *head ){
    struct ListNode *q, *p = head, *tail = NULL;
    while(p){
        q = (struct ListNode*)malloc(sizeof(struct ListNode));
        q->data = p->data;
        if(tail == NULL) q->next = NULL;
        else q->next = tail;
        tail = q;
        p = p->next;
    }
    return tail;
}

6-1 顺序表基本操作

bool ListInsert(SqList *&L,int i,ElemType e){
    if(i <= 0 || i > L->length + 1 || L->length == MaxSize) return false;
    i -- ;
    if(i == L->length){
        L->data[i] = e;
        L->length ++ ;
    }
    else{
        for(int j = L->length; j > i; j -- ) L->data[j] = L->data[j - 1];
        L->data[i] = e;
        L->length ++ ;
        //e = L->data[i - 1];
    }
    return true;
}
bool ListDelete(SqList *&L,int i,ElemType &e){
    if(i < 1 || i > L->length) return false;
    i -- ;
    e = L->data[i];
    for(int j = i; j < L->length; j ++ ) L->data[j] = L->data[j + 1];
    L->length -- ;
    return true;
} 
bool ListDeleteElem(SqList *&L,ElemType e){
    int i = 0, j = 0, flag = 0;
    while(i + j < L->length){
        L->data[i] = L->data[i + j];
        if(L->data[i] == e){
            j ++ ;
            flag = 1;
        }
        else i ++ ;
    }
    L->length = L->length - j;
    if(flag == 0) return false;
    return true;
}

7-1 列车调度

循环遍历就会导致超时

#include<stdio.h>
#include<stdlib.h>
int n, cnt = 1;
int a[100010], t[100010];
int main(){
    scanf("%d", &n);
    for(int i = 0; i < n; i ++ ){
        scanf("%d", &a[i]);
    }
    t[0] = a[0];
    for(int i = 1; i < n; i ++ ){
        for(int j = 0; j < cnt; j ++ ){
            if(t[j] > a[i]){
                t[j] = a[i];
                break;
            }
            else if(j == cnt - 1){
                t[cnt ++ ] = a[i];
                break;
            }
        }
    }
    printf("%d", cnt);
    return 0;
}

所以就需要用二分法来解决；用到了查找精确值的二分法

#include<stdio.h>
#include<stdlib.h>
int n, cnt = 1;

int main(){
    scanf("%d", &n);
    int a[n], t[n];
    for(int i = 0; i < n; i ++ ){
        scanf("%d", &a[i]);
    }
    t[0] = a[0];
    for(int i = 1; i < n; i ++ ){
        if(a[i] > t[cnt - 1]){
            t[cnt ++ ] = a[i];
        }
        else{
            int l = 0, r = cnt - 1;
            while(l < r){
                int mid = l + r >> 1;
                if(t[mid] > a[i]) r = mid - 1;
                else l = mid + 1;
            }
            t[l] = a[i];
        }
    }
    printf("%d", cnt);
    return 0;
}

7-2 愿天下有情人都是失散多年的兄妹

五代人需要用追溯，dfs

关于坑点就是对父母的性别，当父母为可考的时候，要赋上性别，为后面追溯同祖先并查集准备。

if (father != -1) a[father].sex = 'M';
if (mother != -1) a[mother].sex = 'F';

还有就是pta上只能c，c的结构体和c++设置有一点点不一样（

#include<stdio.h>
#define max 100000
typedef struct Node{
    int father;
    int mother;
    char sex;
}node;
node a[max];
int n, k;
int pre[max];
int find(int x){
    if(x != pre[x]) pre[x] = find(pre[x]);
    return pre[x];
}
void Union(int a, int b){
    if (b == -1) return ;
    int pa = find(a);
    int pb = find(b);
    if (pa != pb) pre[pa] = pb;
}
int dfs(int x, int y, int age){
    if (!(x != -1 && y != -1)) return 1;
    if (age > 5) return 1;
    if (x == y) return 0;
    else
        return dfs(a[x].father, a[y].father, age + 1) && dfs(a[x].father, a[y].mother, age + 1) && dfs(a[x].mother, a[y].father, age + 1) && dfs(a[x].mother, a[y].mother, age + 1);
}
int main(){
    for (int i = 0; i < max; i++){
        pre[i] = i;
        a[i].father = -1;
        a[i].mother = -1;
    }
    scanf("%d", &n);
    for (int i = 0; i < n; i++){
        int id, father, mother;
        char sex;
        scanf("%d %c %d %d", &id, &sex, &father, &mother);
        a[id].father = father;
        a[id].sex = sex;
        a[id].mother = mother;
        Union(id, father);
        Union(id, mother);
        if (father != -1) a[father].sex = 'M';
        if (mother != -1) a[mother].sex = 'F';
    }
    scanf("%d", &k);
    for (int i = 0; i < k; i++){
        int x, y;
        scanf("%d %d", &x, &y);
        if (a[x].sex == a[y].sex) printf("Never Mind\n");
        else{
            if (find(x) != find(y)) printf("Yes\n");
            else if (dfs(x, y, 1)) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}

7-3 修理牧场

还不是特别会（

#include<stdio.h>
#include<stdlib.h>
int n;
int a[10010];//原来是一开始开小了，紫砂
int cmp(const void *a, const void *b){
    return *(int*)a - *(int*)b;
}
int main(){
    int sum = 0, pay = 0, j;
    scanf("%d", &n);
    for(int i = 0; i < n; i ++ ){
        scanf("%d", &a[i]);
    }
    qsort(a, n, sizeof(a[0]), cmp);
    for(int i = 0; i < n - 1; i ++ ){
        sum = a[i] + a[i + 1];
        pay = pay + sum;
        for(j = i + 2; j < n; j ++ ){
            if(sum > a[j]) a[j - 1] = a[j];
            else{
                a[j - 1] = sum;
                break;
            }
        }
        if(j == n) a[n - 1] = sum;
    }
    printf("%d", pay);
    return 0;
}