PAT甲级2019冬季考试题解

A Good In C
纯模拟题，用string数组读入数据，注意单词数量的判断

#include<bits/stdc++.h>
using namespace std;
const int maxn=;
string a[][];
int main () {
    for (int i=;i<=;i++) {
        for (int j=;j<=;j++)
        cin>>a[i][j];
    }
    string t;
    getchar ();
    getline (cin,t);
    string s[];
    int top=;
    for (int i=;i<t.length();i++) {
        if (t[i]>='A'&&t[i]<='Z') s[top]+=t[i];
        else if (t[i-]>='A'&&t[i-]<='Z') top++;
    }
    if (top!=&&!(t[t.length()-]>='A'&&t[t.length()-]<='Z')) top--;
    for (int k=;k<=top;k++) {
        for (int j=;j<=;j++) {
            for (int i=;i<s[k].length();i++) {
                if (i!=) printf (" ");
                cout<<a[s[k][i]-'A'+][j];
            }
            cout<<endl;
        }
        //for (int j=1;j<=7;j++)
        if (k!=top) printf ("\n");
    }
    //printf ("\n");
    return ;
}

B Block Reversing
链表常规题，用结构体静态建立链表，有两个注意点：1、给出的结点里会有无关结点。2、链表里的元素个数能不能整除k要分开判断。

#include<bits/stdc++.h>
using namespace std;
const int maxn=;
struct node {
    int address;
    int data;
    int next;
}Node[maxn];
int main () {
    int head,N,K;
    scanf ("%d %d %d",&head,&N,&K);
    int address,data,next;
    for (int i=;i<N;i++) {
        scanf ("%05d %d %05d",&address,&data,&next);
        Node[address].address=address;
        Node[address].data=data;
        Node[address].next=next;
    }
    vector<node> vi;
    while (head!=-) {
        vi.push_back(Node[head]);
        head=Node[head].next;
    }
    vector<int> vp[];
    if (vi.size()%K!=) {
        printf ("%05d %d ",vi[vi.size()/K*K].address,vi[vi.size()/K*K].data);
        for (int i=vi.size()/K*K+;i<vi.size();i++)
        printf ("%05d\n%05d %d ",vi[i].address,vi[i].address,vi[i].data);
        for (int i=vi.size()/K;i>;i--) {
            for (int j=(i-)*K;j<i*K;j++)
            printf ("%05d\n%05d %d ",vi[j].address,vi[j].address,vi[j].data);
        }
        printf ("-1");
    }
    else {
        printf ("%05d %d ",vi[vi.size()-K].address,vi[vi.size()-K].data);
        for (int i=vi.size()/K;i>;i--) {
            int flag=;
            if (i==vi.size()/K) flag=;
            for (int j=(i-)*K+flag;j<i*K;j++)
            printf ("%05d\n%05d %d ",vi[j].address,vi[j].address,vi[j].data);
        }
        printf ("-1");
    }
    return ;
} 

C Summit
个人认为全场最简单的一题，也是第一道ac的题目，就是图论，根据题意暴力做即可~

#include<bits/stdc++.h>
using namespace std;
const int maxn=;
const int inf=1e9;
int g[maxn][maxn];
int N,M,K,L,a[maxn];
void init () {
    for (int i=;i<=N;i++)
    for (int j=;j<=N;j++)
    g[i][j]=inf;
}
int main () {
    scanf ("%d %d",&N,&M);
    int u,v;
    init ();
    for (int i=;i<M;i++) {
        scanf ("%d %d",&u,&v);
        g[u][v]=g[v][u]=;
    }
    scanf ("%d",&K);
    for (int i=;i<K;i++) {
        int flag1=,flag2=,hash1[maxn]={};
        scanf ("%d",&L);
        for (int j=;j<L;j++) {
            scanf ("%d",&a[j]);
            hash1[a[j]]=;
        }
        for (int j=;j<L;j++) {
            for (int k=j+;k<L;k++) {
                if (g[a[j]][a[k]]==inf) {
                    flag1=;break;
                }
            }
            if (flag1==) break;
        }
        if (flag1==) {
            printf ("Area %d needs help.\n",i+);
            continue;
        }
        int u;
        for (int j=;j<=N;j++) {
            if (hash1[j]==) {
                int flag3=;
                for (int k=;k<L;k++)
                if (g[a[k]][j]==inf) flag3++;
                if (flag3==) {
                    u=j;
                    flag2=;
                    break;
                }
            }
            if (flag2==) break;
        }
        if (flag1==&&flag2==) {
            printf ("Area %d may invite more people, such as %d.\n",i+,u);
        }
        else if (flag1==&&flag2==) {
            printf ("Area %d is OK.\n",i+);
        }
    }
    return ;
}

D Cartesian Tree
一开始以为是堆的算法，写了很久，其实并不是。一颗树，结点数据符合小顶堆的规则，给出它的中序遍历，要求输出层序。解决方法是先找出给定序列里最小的点作为根结点，再以根节点为标志，找根节点左边和右边各自最小的结点，作为根节点的左右子树，递归处理。

#include<bits/stdc++.h>
using namespace std;
const int maxn=;
struct node {
    int data;
    node * left;
    node * right;
};
int a[maxn];
node * create (int inL,int inR,int inRoot) {
    if (inRoot==1e9) return NULL;
    node * root=new node;
    root->data=a[inRoot];
    int inLeft=1e9,inRight=1e9;
    int min=1e9;
    for (int i=inL;i<inRoot;i++) {
        if (a[i]<min) {
            min=a[i];
            inLeft=i;
        }
    }
    root->left=create (inL,inRoot-,inLeft);
    min=1e9;
    for (int i=inRoot+;i<=inR;i++) {
        if (a[i]<min) {
            min=a[i];
            inRight=i;
        }
    }
    root->right=create (inRoot+,inR,inRight);
    return root;
}
void bfs (node * root) {
    queue<node *> q;
    q.push(root);
    while (!q.empty()) {
        node * now=q.front();
        q.pop();
        if (now!=root) printf (" ");
        printf ("%d",now->data);
        if (now->left) q.push(now->left);
        if (now->right) q.push(now->right);
    }
}
int main () {
    int N;
    scanf ("%d",&N);
    for (int i=;i<=N;i++) {
        scanf ("%d",&a[i]);
    }
    int min=1e9,inRoot=-;
    for (int i=;i<=N;i++) {
        if (a[i]<min) {
            min=a[i];
            inRoot=i;
        }
    }
    //printf ("%d\n",inRoot);
    node * root=create (,N,inRoot);
    bfs (root);
    return ;
}