PAT（甲级）2020年秋季考试
1. 7-1 Panda and PP Milk (20 分)
这个题目刚开始想的是先找出最轻的熊猫，然后和两边的熊猫体重比较，重则喝的牛奶加100，否则不变。然后按照这个方法对熊猫按体重由小到大来遍历，结果有3 5两个测试点过不去,不知道错在哪里。
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node{
int weight, pos;
};
bool cmp(node a, node b)
{
if(a.weight != b.weight)
return a.weight < b.weight;
}
int main()
{
vector<node> v;
int n;
scanf("%d", &n);
vector<int> w(n), m(n, 0);
for(int i = 0; i < n; i++)
{
scanf("%d", &w[i]);
v.push_back({
w[i], i});
}
sort(v.begin(), v.end(), cmp);
for(int i = 0; i < n; i++)
{
int pos = v[i].pos;
if(m[pos] == 0)
m[pos] = 200;
if(pos+1 < n)
{
if(w[pos] < w[pos+1])
m[pos+1] = m[pos]+100;
else if(w[pos] == w[pos+1])
m[pos+1] = m[pos];
}
if(pos-1 >= 0)
{
if(w[pos] < w[pos-1])
m[pos-1] = m[pos]+100;
else if(w[pos] == w[pos-1])
m[pos-1] = m[pos];
}
}
int total = 0;
for(int i = 0; i < n; i++)
total += m[i];
printf("%d", total);
return 0;
}
之后看了下这个博主的解析，方法更加简便 ->更简便的方法
2.7-2 How Many Ways to Buy a Piece of Land (25 分)(AC)
这道题目没有多想，直接用的暴力破解：
#include<cstdio>
#include<vector>
using namespace std;
vector<int> v;
int n, m, cnt = 0;
int main()
{
scanf("%d%d", &n, &m);
v.resize(n);
for(int i = 0; i < n; i++)
scanf("%d", &v[i]);
for(int i = 0; i < n; i++)
{
int total = v[i];
if(total <= m)
{
cnt++;
for(int j = i+1; j < n; j++)
{
if(total + v[j] <= m)
{
cnt++;
total += v[j];
}
else
break;
}
}
}
printf("%d", cnt);
}
后来看了一个博主更加简便的做法：用sum[i]表示前i个和相加，然后用sum[i]与从k开始的sum作差，如果此时的差大于所给的数说明钱不够买从k+1到 i 的岛，于是与sum[k+1]作差，直至找到一个小于所给数的j，即所给的钱能买从j开始到i的连续岛。而j-i表示可以选择的方式，sum[j+1]表示只买j+1,sum[j+2]表示买j+1~j+2…->博主链接
#include <iostream>
using namespace std;
const int N = 10010;
int n,m;
int sum[N];
int main(){
cin >> n >> m;
for(int i=1;i<=n;i++){
int x;cin >> x;
sum[i] = x + sum[i - 1];
}
int res = 0;
for(int i=1,j=0;i<=n;i++)
{
while(sum[i] - sum[j] > m)
j ++ ;
res += i - j;//因为每一个sum代表前i项和，所以i-j表示可以选j+1, j+1~j+2,j+1~j+3......j+1~i共i-j个选择
}
cout << res << endl;
return 0;
}
3.7-3 Left-View of Binary Tree (25 分)（AC)
模板题：先建树，后层次遍历，把每层遍历的第一个存储后输出
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
vector<int> pre, in, depth[20];
struct node{
int data, depth;
node * left, * right;
};
node * root;
void create(node * &root, int prel, int prer, int inl, int inr)
{
if(prel > prer)
return;
root = new node;
root->data = pre[prel];
root->left = root->right = NULL;
int i = inl;
while(in[i] != root->data)
i++;
int leftNum = i-inl;
create(root->left, prel+1, prel+leftNum, inl, i-1);
create(root->right, prel+leftNum+1, prer, i+1, inr);
}
void BFS(node * root)
{
queue<node * > q;
q.push(root);
while(!q.empty())
{
node * front = q.front();
q.pop();
if(front->left != NULL)
{
q.push(front->left);
front->left->depth = front->depth+1;
depth[front->left->depth].push_back(front->left->data);
}
if(front->right != NULL)
{
q.push(front->right);
front->right->depth = front->depth+1;
depth[front->right->depth].push_back(front->right->data);
}//保存每层的第一个结点
}
}
int main()
{
int n;
scanf("%d", &n);
pre.resize(n+1), in.resize(n+1);
for(int i = 1; i <= n; i++)
scanf("%d", &in[i]);
for(int i = 1; i <= n; i++)
scanf("%d", &pre[i]);
create(root, 1, n, 1, n);
root->depth = 1;
depth[1].push_back(root->data);
BFS(root);
for(int i = 1; i < 20; i++)
{
if(depth[i].size() > 0)
{
if(i > 1)
printf(" ");
printf("%d", depth[i][0]);
}
else break;
}
}
4.7-4 Professional Ability Test (30 分)
这个题目又臭又长，导致到最后自己都不知道在写些什么东西了，题目也没有完全读懂，只是用测试样例判断okay是不出现环，impossible是存在环这个样子，所以用了BFS来判断有没有环。找所有入度为0的点入队，然后出队的同时把相连的点的入度-1再入队，这样到最后如果出队的点有n个表示能够构成拓扑，否则存在环。至于找最短的路径，则是用的DFS算法来找到最小s与最大d然后把途中经过的路径保存在ans中。结果最后一个测试点超时扣了3分。
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
int n, m;
const int maxn = 1010;
vector<int> ind, ind2, que, link[maxn], temp, ans;
int score[maxn][maxn], value[maxn][maxn];
bool vis[maxn];
int mins, maxd;
queue<int> q;
bool BFS( )
{
int cnt = 0;
while(q.size() > 0)
q.pop();
for(int i = 0; i < n; i++)
if(ind2[i] == 0)
{
q.push(i);
vis[i] = true;
}
while(!q.empty())
{
int front = q.front();
cnt++;
q.pop();
for(int i = 0; i < link[front].size(); i++)
{
int id = link[front][i];
ind2[id]--;
if(vis[id] == false && ind2[id] == 0)
{
q.push(id);
vis[id] = true;
}
}
}
if(cnt == n)
return true;
else
return false;
}
void DFS(int start, int des, int totals, int totald)
{
if(start == des && totals > 0 && totald > 0)//防止刚开始进入就start = des而此时totals = 0, totald = 0导致跳出函数
{
if(totals < mins)
{
ans = temp;
mins = totals;
maxd = totald;
}
else if(totals == mins && totald > maxd)
{
ans = temp;
maxd = totald;
}
return;
}
else if(totals > mins) return;
for(int i = 0; i < link[start].size(); i++)
{
int id = link[start][i];
temp.push_back(id);
DFS(id, des, totals+score[start][id], totald+value[start][id]);
temp.pop_back();
}
}
int main()
{
scanf("%d%d", &n, &m);
ind.resize(n, 0);
for(int i = 0; i < m; i++)
{
int a, b;
scanf("%d%d", &a, &b);
scanf("%d%d", &score[a][b], &value[a][b]);
ind[b]++;
link[a].push_back(b);
}
ind2 = ind;
int query, num;
scanf("%d", &query);
que.resize(query);
int flag = 0;
for(int i = 0; i < query; i++)
scanf("%d", &que[i]);
if(BFS())
{
printf("Okay.");
flag = 1;
}
else
printf("Impossible.");
for(int i = 0; i < query; i++)
{
int id = que[i];
if(ind[id] == 0)
printf("\nYou may take test %d directly.", id);
else if(flag == 1)
{
printf("\n");
mins = 1000000000, maxd = -1;
ans.clear(), temp.clear();
for(int j = 0; j < n; j++)
if(ind[j] == 0 &&j != id)
{
temp.push_back(j);
DFS(j, id, 0, 0);
temp.pop_back();
}
for(int k = 0; k < ans.size(); k++)
{
printf("%d", ans[k]);
if(k < ans.size()-1) printf("->");
}
}
else if(flag == 0)
printf("\nError.");
}
}