PAT（甲级）2017年秋季考试

PAT（甲级）2017年秋季考试

D题红黑树待补21/30
大佬的代码，看着想哭，这才是艺术啊

A Cut Integer

模拟题

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int n;

int getLen(ll x){
	int len = 0;
	while(x){
		len++;
		x/=10;
	}
	return len;
}

int main(){
	cin>>n;
	while(n--){
		ll x;
		cin>>x;
		int len = getLen(x);
		ll temp = x;
		int t = 0;
		ll right = 0;
		ll rightTemp = 0;
		ll left = 0;
		while(t<len/2){
			rightTemp = rightTemp * 10 + temp%10;
			temp/=10;
			t++;
		}
		while(rightTemp){
			right = right * 10 + rightTemp%10;
			rightTemp/=10;
		}
		left = temp;
		if(left == 0 || right == 0){
			puts("No");
		}else{
			if(x%(left*right) == 0) puts("Yes");
			else puts("No");
		}
	}
	return 0;
} 

B Splitting A Linked List

链表题

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1010;
const int maxm = 100000;
struct node{
	int address;
	int data;
	int next;
}nod[maxm];

vector<node> vec;
vector<node> ans;
int Head,n,k;
int vis[maxm];

int main(){
	cin>>Head>>n>>k;
	for(int i=1;i<=n;i++){
		int add,dat,nex;
		cin>>add>>dat>>nex;
		nod[add].address = add;
		nod[add].data = dat;
		nod[add].next = nex;
	}
	for(int head = Head;head!=-1;head=nod[head].next){
		vec.push_back(nod[head]);
	}
	for(int i=0;i<vec.size();i++){
		if(vec[i].data < 0 && !vis[vec[i].address]){
			ans.push_back(vec[i]);
			vis[vec[i].address] = 1;
		}
	}
	for(int i=0;i<vec.size();i++){
		if(vec[i].data >= 0 && vec[i].data <= k && !vis[vec[i].address]){
			ans.push_back(vec[i]);
			vis[vec[i].address] = 1;
		}
	}
	for(int i=0;i<vec.size();i++){
		if(!vis[vec[i].address]){
			ans.push_back(vec[i]);
			vis[vec[i].address] = 1;
		}
	}
	if(ans.size() == 1){
		printf("%05d %d -1",ans[0].address,ans[0].data);
		return 0;
	}
	for(int i=0;i<ans.size()-1;i++){
		printf("%05d %d %05d
",ans[i].address,ans[i].data,ans[i+1].address);
	}
	if(ans.size() > 1) printf("%05d %d -1",ans[ans.size()-1].address,ans[ans.size()-1].data);
	return 0;
}

C Vertex Cover

简单图论，最小覆盖，邻接表存图

#include<bits/stdc++.h>
using namespace std;

const int maxn = 10000;
int n,m,k;
int g[maxn][maxn];
int vis[maxn][maxn];
vector<int> vec;
void init(){
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++){
			vis[i][j] = 0;
		}
	}
}

int main(){
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int u,v;
		cin>>u>>v;
		g[u][v] = 1;
		g[v][u] = 1;
	}
	cin>>k;
	for(int t=0;t<k;t++){
		init();
		int nv;
		cin>>nv;
		for(int i=0;i<nv;i++) {
			int d;
			cin>>d;
			vec.push_back(d);
		}
		for(int i=0;i<=nv-1;i++){
			for(int j=0;j<n;j++){
				if(g[vec[i]][j] == 1){
					vis[vec[i]][j] = 1;
					vis[j][vec[i]] = 1;
				}
			}
		}
		bool flag = true;
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				if(g[i][j] == 1 && (vis[i][j] == 0 || vis[j][i] == 0)){
					flag = false;
					break;
				}
			}
			if(flag == false) break;
		}
		if(flag) puts("Yes");
		else puts("No");
		vec.clear();
	}
	return 0;
}

D Is It A Red-Black Tree

判断是否红黑树 21分/30分 待补

#include<bits/stdc++.h>
using namespace std;

/*
21分 
*/

const int maxn = 1010;
int k,n;
int pre[maxn];
int rb[maxn];
int xb[maxn];
//bool childFlag = true;
struct node{
	int v;
	int cnt;
	node *l;
	node *r;
};

void insert(node *root,int pos){
	if(root == NULL) return;
	if(rb[pos] == 0){
		root->cnt++;
	}
	if(root->v > pre[pos]){
		if(root->l == NULL){
			root->l = new node();
			root->l->v = pre[pos];
			root->l->l = NULL;
			root->l->r = NULL;
			if(rb[pos] == 0) root->l->cnt = 1;
		}else{
			insert(root->l,pos);
		}
	}else{
		if(root->r == NULL){
			root->r = new node();
			root->r->v = pre[pos];
			root->r->l = NULL;
			root->r->r = NULL;
			if(rb[pos] == 0) root->r->cnt = 1;
		}else{
			insert(root->r,pos);
		}	
	}
}

bool checkChild(node *Root){
	if(Root->l){
		if(rb[Root->v] < 0 && rb[Root->l->v] <0){
			return false;
		}else{
			return checkChild(Root->l);
		}
	}
	if(Root->r){
		if(rb[Root->v] < 0 && rb[Root->r->v] <0){
			return false;
		}else{
			return checkChild(Root->r);
		}
	}
	return true;
}

void init(){
	for(int i=0;i<=n;i++){
		rb[i] = 0;
		xb[i] = 0;
		pre[i] = 0;
	}
}

void bfs(node *root){
	if(root == NULL) return;
	cout<<root->v<<" "<<root->cnt<<endl; 
	if(root->l) bfs(root->l);
	if(root->r) bfs(root->r);
}

bool can = true;
void travel(node *root){
	if(can == false) return;
	if(root->v == 15){
		int ddd = 1;
	}
	if(root->l && root->r){
		if(root->l->cnt != root->r->cnt) {
			can = false;
			return;
		}
	}
	if(root->l) {
		travel(root->l);
	}
	if(root->r) {
		travel(root->r);
	}
}

int main(){
	cin>>k;
	while(k--){
		cin>>n;
		init();
		for(int i=1;i<=n;i++) {
			int d;
			cin>>d;
			if(d < 0 ){
				xb[-d] = 1;
				rb[i] = 1;
				pre[i] = -d;
			}else{
				rb[i] = 0;
				pre[i] = d;
			}
		}
		node *Root = new node(); 
		Root->l = NULL;
		Root->r = NULL;
		Root->v = pre[1];
		Root->cnt = 1;
		for(int i=2;i<=n;i++) insert(Root,i);
		if(rb[1] == 1) puts("No");
		else if(checkChild(Root) == false) puts("No");
		else{
			//bfs(Root);
			can = true;
			travel(Root);
			if(can == false) puts("No");
			else puts("Yes");
		}
	}
	return 0;
}