7-1 Panda and PP Milk (20 分)
时间限制：150 ms 内存限制：64 MB
PP milk （盆盆奶）is Pandas' favorite. They would line up to enjoy it as show in the picture. On the other hand, they could drink in peace only if they believe that the amount of PP milk is fairly distributed, that is, fatter panda can have more milk, and the ones with equal weight may have the same amount. Since they are lined up, each panda can only compare with its neighbor(s), and if it thinks this is unfair, the panda would fight with its neighbor.
Given that the minimum amount of milk a panda must drink is 200 ml. It is only when another bowl of milk is at least 100 ml more than its own that a panda can sense the difference.
Now given the weights of a line of pandas, your job is to help the breeder（饲养员）to decide the minimum total amount of milk that he/she must prepare, provided that the pandas are lined up in the given order.
Input Specification:
Each input file contains one test case. For each case, first a positive integer n (≤104) is given as the number of pandas. Then in the next line, n positive integers are given as the weights (in kg) of the pandas, each no more than 200. the numbers are separated by spaces.
Output Specification:
For each test case, print in a line the minimum total amount of milk that the breeder must prepare, to make sure that all the pandas can drink in peace.
Sample Input:
10
180 160 100 150 145 142 138 138 138 140
Sample Output:
3000
Hint:
The distribution of milk is the following:
400 300 200 500 400 300 200 200 200 300
解析：
“臭名昭著”的难题……
方法一：每个熊猫要喝的奶只与它左右两侧的熊猫有关。PAT题库里有两道类似的题：【B1045/A1101 Quick Sort】【B1040/A1093 Count PAT's】。前者要求判断数组上每个值左边所有数是否都比该值小，以及右边所有数是否都比该值大。后者可以转化为以字母A为参照，统计字符串中每个A字符左边P字符的个数，以及右边T字符的个数。
这道题也可以用类似的思路，扫两遍数组，第一遍从左往右分配，当前熊猫体重大于左边就+100，等于就分一样的奶，小于左边就只分配200，这样每个熊猫不会和左边的熊猫打架；第二遍从右往左分配，方法相同，这样每个熊猫不会和右边的熊猫打架。最后把两个数组合并，合并方式是取较大值。
#include <vector>
#include <iostream>
using namespace std;
void test () {
int n, i, sum = 0;
scanf("%d", &n);
vector<int> input(n), a(n), b(n);
for (i = 0; i < n; i++) scanf("%d", &input[i]);
a[0] = 200;
for (i = 1; i < n; i++) {
if (input[i] > input[i - 1]) a[i] = a[i - 1] + 100;
else if (input[i] == input[i - 1]) a[i] = a[i - 1];
else a[i] = 200;
}
b.back() = 200;
for (i = n - 2; i >= 0; i--) {
if (input[i] > input[i + 1]) b[i] = b[i + 1] + 100;
else if (input[i] == input[i + 1]) b[i] = b[i + 1];
else b[i] = 200;
sum += max(a[i], b[i]);
}
sum += max(a.back(), b.back());
printf("%d", sum);
}
int main() {
test();
return 0;
}
方法二：如果体重从前往后不变/递增，那么分配的奶会相同/变多。如果下降，要分配的奶不好确定，因为要受到右边熊猫的影响，但是可以先分配200。 接着判断左边的熊猫是否满意，如果不满意，就给左边的熊猫+100，接着继续判断左边的左边的熊猫是否满意，直到某个左边的熊猫满意为止。
#include <vector>
#include <iostream>
using namespace std;
int main(){
int n, sum = 0, i, j;
scanf("%d", &n);
vector<int> panda(n), milk(n);
scanf("%d", &panda[0]);
milk[0] = 200;
for(i = 1;i < n; i++){
scanf("%d", &panda[i]);
if (panda[i] == panda[i - 1]) milk[i] = milk[i - 1];
else if (panda[i] > panda[i - 1]) {
milk[i] = milk[i - 1] + 100;
} else {
milk[i] = 200;
for(j = i - 1; j >= 0; j--) {
if (panda[j] >= panda[j + 1] && milk[j] <= milk[j + 1])
milk[j] += 100;
else break;
}
}
}
for(i = 0; i < n; i++){
sum += milk[i];
}
printf("%d", sum);
return 0;
}
7-2 How Many Ways to Buy a Piece of Land (25 分)
时间限制：150 ms 内存限制：64 MB
The land is for sale in CyberCity, and is divided into several pieces. Here it is assumed that each piece of land has exactly two neighboring pieces, except the first and the last that have only one. One can buy several contiguous（连续的） pieces at a time. Now given the list of prices of the land pieces, your job is to tell a customer in how many different ways that he/she can buy with a certain amount of money.
Input Specification:
Each input file contains one test case. Each case first gives in a line two positive integers: N (≤104), the number of pieces of the land (hence the land pieces are numbered from 1 to N in order), and M (≤109), the amount of money that your customer has.
Then in the next line, N positive integers are given, where the i-th one is the price of the i-th piece of the land.
It is guaranteed that the total price of the land is no more than 109.
Output Specification:
For each test case, print the number of different ways that your customer can buy. Notice that the pieces must be contiguous.
Sample Input:
5 85
38 42 15 24 9
Sample Output:
11
Hint:
The 11 different ways are:
38
42
15
24
9
38 42
42 15
42 15 24
15 24
15 24 9
24 9
解析：
题目大意：有一组正整数，求连续的子序列，使子序列的和 ≤ 给定的数，问这样的子序列有多少个。
方法一：DFS+剪枝，来得及。
#include <vector>
#include <iostream>
using namespace std;
int n, m;
vector<int> prices, ans;
int count = 0;
void DFS (int index, int sum) {
if (index >= n || sum > m) return;
if (sum + prices[index] <= m) {
ans.push_back(prices[index]);
count++;
DFS(index + 1, sum + prices[index]);
}
}
void test() {
int i, j, temp;
scanf("%d %d", &n, &m);
for (i = 0; i < n; i++) {
scanf("%d", &temp);
prices.push_back(temp);
}
for (i = 0; i < n; i++) {
ans.clear();
DFS(i, 0);
}
printf("%d", count);
}
int main() {
test();
return 0;
}
方法二：既然是连续的子序列的和，可以求前缀和，然后双重循环暴力找。
优化：前缀和一定是有序的，内层循环可以用二分找到第一个大于sums[i]＋M的数，而不是顺序查找，就不贴代码啦~（其实是懒得再花5块钱验证了……）
#include <vector>
#include <iostream>
using namespace std;
int N, M;
vector<int> sums;
int ans = 0;
void test () {
int i, j, temp1, sum = 0;
scanf("%d %d", &N, &M);
sums.push_back(0);
for (i = 0; i < N; i++) {
scanf("%d", &temp1);
sum += temp1;
sums.push_back(sum);
}
for (i = 0; i < (int)sums.size() - 1; i++) {
for (j = i + 1; j < sums.size(); j++) {
if (sums[j] - sums[i] <= M) ans++;
}
}
printf("%d", ans);
}
int main() {
test();
return 0;
}
7-3 Left-View of Binary Tree (25 分)
时间限制：400 ms 内存限制：64 MB
The left-view of a binary tree is a list of nodes obtained by looking at the tree from left hand side and from top down. For example, given a tree shown by the figure, its left-view is { 1, 2, 3, 4, 5 }
Given the inorder and preorder traversal sequences of a binary tree, you are supposed to output its left-view.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤20), which is the total number of nodes in the tree. Then given in the following 2 lines are the inorder and preorder traversal sequences of the tree, respectively. All the keys in the tree are distinct positive integers in the range of int.
Output Specification:
For each case, print in a line the left-view of the tree. All the numbers in a line are separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.
Sample Input:
8
2 3 1 5 4 7 8 6
1 2 3 6 7 4 5 8
Sample Output:
1 2 3 4 5
解析：
方法一：根据中序和后序建树，然后层次遍历，每层只记录第一个点。
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
typedef struct Node {
int data, layer;
struct Node *l = NULL, *r = NULL;
} Node;
Node *root = NULL;
int N;
vector<int> pre, in, ans;
Node *create (int pre1, int pre2, int in1, int in2) {
if (pre1 > pre2) return NULL;
int i = in1;
while (in[i] != pre[pre1]) i++;
int len = i - in1;
Node *node = new Node;
node->data = pre[pre1];
node->l = create(pre1 + 1, pre1 + len, in1, i - 1);
node->r = create(pre1 + len + 1, pre2, i + 1, in2);
return node;
}
void BFS () {
queue<Node*> q;
q.push(root);
root->layer = 1;
ans.push_back(root->data);
while (!q.empty()) {
Node *node = q.front(); q.pop();
// 每层只记录第一个点
if (node->l != NULL) {
node->l->layer = node->layer + 1;
q.push(node->l);
if (ans.size() == node->layer) ans.push_back(node->l->data);
}
if (node->r != NULL) {
node->r->layer = node->layer + 1;
q.push(node->r);
if (ans.size() == node->layer) ans.push_back(node->r->data);
}
}
}
void test () {
int i;
scanf("%d", &N);
pre.resize(N); in.resize(N);
for (i = 0; i < N; i++) scanf("%d", &in[i]);
for (i = 0; i < N; i++) scanf("%d", &pre[i]);
root = create(0, N - 1, 0, N - 1);
BFS();
for (i = 0; i < ans.size(); i++) {
printf("%d", ans[i]);
if (i < ans.size() - 1) printf(" ");
}
}
int main() {
test();
return 0;
}
方法二：不用建树。
#include <vector>
#include <iostream>
using namespace std;
int n;
vector<int> pre, in, ans;
int max_layer = -1;
void create (int left1, int right1, int left2, int right2, int layer) {
if (left1 > right1) return;
int node = pre[left1], i = left2;
while (in[i] != node) i++;
// 因为是先处理左子树再处理右子树，该层处理的第一个结点一定符合要求
if (ans[layer] == 0) ans[layer] = node;
if (max_layer < layer) max_layer = layer;
int L = i - left2;
create(left1 + 1, left1 + L, left2, i - 1, layer + 1);
create(left1 + 1 + L, right1, i + 1, right2, layer + 1);
}
void test() {
int i;
scanf("%d", &n);
pre.resize(n); in.resize(n);
ans.resize(n, 0); // ans最长为n
for (i = 0; i < n; i++) scanf("%d", &in[i]);
for (i = 0; i < n; i++) scanf("%d", &pre[i]);
create(0, n - 1, 0, n - 1, 0);
for (i = 0; i <= max_layer; i++) {
printf("%d", ans[i]);
if (i < max_layer) printf(" ");
}
}
int main(){
test();
return 0;
}
7-4 Professional Ability Test (30 分)
时间限制：3000 ms 内存限制：64 MB
Professional Ability Test (PAT) consists of several series of subject tests. Each test is divided into several levels. Level A is a prerequisite (前置要求) of Level B if one must pass Level A with a score no less than S in order to be qualified to take Level B. At the mean time, one who passes Level A with a score no less than S will receive a voucher（代金券）of D yuans (Chinese dollar) for taking Level B.
At the moment, this PAT is only in design and hence people would make up different plans. A plan is NOT consistent if there exists some test T so that T is a prerequisite of itself. Your job is to test each plan and tell if it is a consistent one, and at the mean time, find the easiest way (with minimum total S) to obtain the certificate of any subject test. If the easiest way is not unique, find the one that one can win the maximum total value of vouchers.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (≤1000) and M, being the total numbers of tests and prerequisite relations, respectively. Then M lines follow, each describes a prerequisite relation in the following format:
T1 T2 S D
where T1 and T2 are the indices (from 0 to N−1) of the two distinct tests; S is the minimum score (in the range (0, 100]) required to pass T1 in order to be qualified to take T2; and D is the value of the voucher (in the range (0, 500]) one can receive if one passes T1 with a score no less than S and plan to take T2. It is guaranteed that at most one pair of S and D are defined for a prerequisite relation.
Then another positive integer K (≤N) is given, followed by K queries of tests. All the numbers in a line are separated by spaces.
Output Specification:
Print in the first line Okay. if the whole plan is consistent, or Impossible. if not.
If the plan is consistent, for each query of test T, print in a line the easiest way to obtain the certificate of this test, in the format:
T0->T1->...->T
However, if T is the first level of some subject test (with no prerequisite), print You may take test T directly. instead.
If the plan is impossible, for each query of test T, check if one can take it directly or not. If the answer is yes, print in a line You may take test T directly.; or print Error. instead.
Sample Input 1:
8 15
0 1 50 50
1 2 20 20
3 4 90 90
3 7 90 80
4 5 20 20
7 5 10 10
5 6 10 10
0 4 80 60
3 1 50 45
1 4 30 20
1 5 50 20
2 4 10 10
7 2 10 30
2 5 30 20
2 6 40 60
8
0 1 2 3 4 5 6 7
Sample Output 1:
Okay.
You may take test 0 directly.
0->1
0->1->2
You may take test 3 directly.
0->1->2->4
0->1->2->4->5
0->1->2->6
3->7
Sample Input 2:
4 5
0 1 1 10
1 2 2 10
3 0 4 10
3 2 5 10
2 0 3 10
2
3 1
Sample Output 2:
Impossible.
You may take test 3 directly.
Error.
解析：
先判断是否DAG，再求路径。DAG的判断算法笔记上已经有模板，照抄即可；而对于求路径，由于起点可能有多个而终点只有一个，如果正向求解，需要遍历每个入度为0的点，容易导致测试点7超时。因此可以采用两种变通的方法：1、倒过来，从唯一的终点出发往回DFS，那么就要建立逆邻接表。2、新建一个顶点作为所有起点的源点，一遍迪杰斯特拉算法即可。
方法一：逆序DFS。
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
const int maxn = 1002, INF = 0x3fffffff;
typedef struct Node {
int next, dist, voucher;
Node (int n, int d, int v) : next(n), dist(d), voucher(v) {}
} Node;
vector<Node> G[maxn], G2[maxn];
int degree[maxn] = {0}, visited[maxn];
int N, M, K;
vector<int> route, ans;
int min_dist, max_voucher;
bool DAG () {
int i, count = 0;
queue<int> q;
for (i = 0; i < N; i++) {
if (degree[i] == 0) {
q.push(i);
}
}
while (!q.empty()) {
int index = q.front(); q.pop();
count++;
for (i = 0; i < G[index].size(); i++) {
int next = G[index][i].next;
degree[next]--;
if (degree[next] == 0) {
q.push(next);
}
}
}
return count == N;
}
void DFS (int index, int dist, int voucher) {
int i;
if (G2[index].empty()){
if (min_dist > dist) {
min_dist = dist; max_voucher = voucher;
route.push_back(index);
ans = route;
route.pop_back();
} else if (min_dist == dist && max_voucher < voucher) {
max_voucher = voucher;
route.push_back(index);
ans = route;
route.pop_back();
}
return;
}
route.push_back(index);
visited[index] = 1;
for (i = 0; i < G2[index].size(); i++) {
Node node = G2[index][i];
if (visited[node.next]) continue;
int new_dist = dist + node.dist, new_voucher = voucher + node.voucher;
if (new_dist <= min_dist) {
DFS(node.next, new_dist, new_voucher);
}
}
visited[index] = 0;
route.pop_back();
}
void test() {
int i, j, t1, t2, t3, t4;
scanf("%d %d", &N, &M);
for (i = 0; i < M; i++) {
scanf("%d %d %d %d", &t1, &t2, &t3, &t4);
G[t1].push_back(Node(t2, t3, t4));
G2[t2].push_back(Node(t1, t3, t4));
degree[t2]++;
}
scanf("%d", &K);
bool flag = DAG();
if (flag) printf("Okay.
");
else printf("Impossible.
");
for (i = 0; i < K; i++) {
scanf("%d", &t1);
if (G2[t1].empty()) {
printf("You may take test %d directly.
", t1);
continue;
}
if (!flag) printf("Error.
");
else {
min_dist = INF, max_voucher = -1;
DFS(t1, 0, 0);
for (j = (int)ans.size() - 1; j >= 0; j--) {
printf("%d", ans[j]);
if (j) printf("->");
}
printf("
");
}
}
}
int main(){
test();
return 0;
}
方法二：一遍迪杰斯特拉。
#include <vector>
#include <unordered_set>
#include <queue>
#include <iostream>
using namespace std;
const int maxn = 1002, INF = 0x3fffffff;
typedef struct Node {
int to, dis1, dis2;
Node (int t, int d, int d2) : to(t), dis1(d), dis2(d2) {}
} Node;
int N, M;
vector<Node> G1[maxn];
int visited[maxn] = {0}, ins[maxn] = {0}, dist[maxn], pre[maxn], price[maxn];
unordered_set<int> starts;
bool DAG () {
queue<int> q;
for (auto item : starts) {
q.push(item);
}
int count = q.size();
while (!q.empty()) {
int index = q.front(); q.pop();
for (auto& item : G1[index]) {
ins[item.to]--;
if (ins[item.to] == 0) {
q.push(item.to); count++;
}
}
}
return count == N;
}
void dij () {
int i, j;
fill(dist, dist + N + 1, INF);
dist[N] = 0;
for (i = 0; i <= N; i++) {
int cur = -1, min_dist = INF;
for (j = 0; j <= N; j++) {
if (visited[j] == 0 && dist[j] < min_dist) min_dist = dist[j], cur = j;
}
visited[cur] = 1;
for (auto& item : G1[cur]) {
if (visited[item.to]) continue;
if (dist[cur] + item.dis1 < dist[item.to]) {
dist[item.to] = dist[cur] + item.dis1;
price[item.to] = price[cur] + item.dis2;
pre[item.to] = cur;
} else if (dist[item.to] == dist[cur] + item.dis1 && price[item.to] < price[cur] + item.dis2) {
price[item.to] = price[cur] + item.dis2;
pre[item.to] = cur;
}
}
}
}
void test () {
int t1, t2, s, d, K;
int i, j;
scanf("%d %d", &N, &M);
for (i = 0; i < M; i++) {
scanf("%d %d %d %d", &t1, &t2, &s, &d);
G1[t1].push_back(Node(t2, s, d));
ins[t2]++;
}
scanf("%d", &K);
for (i = 0; i < N; i++) {
if (ins[i] == 0) starts.insert(i);
}
if (DAG()) {
printf("Okay.
");
for (auto item : starts) {
G1[N].push_back(Node(item, 0, 0));
}
dij();
for (i = 0; i < K; i++) {
scanf("%d", &t1);
if (starts.count(t1)) printf("You may take test %d directly.
", t1);
else {
vector<int> ans;
j = pre[t1];
while (j != N) {
ans.push_back(j);
j = pre[j];
}
for (j = ans.size() - 1; j >= 0; j--) {
printf("%d->", ans[j]);
}
printf("%d
", t1);
}
}
} else {
printf("Impossible.
");
for (i = 0; i < K; i++) {
scanf("%d", &t1);
if (starts.count(t1)) printf("You may take test %d directly.
", t1);
else printf("Error.
");
}
}
}
int main() {
test();
return 0;
}
总结
| 编号 | 标题 | 分数 | 类型 |
|---|---|---|---|
| 7-1 | Panda and PP Milk | 20 | 4.7 其他高效技巧与算法 |
| 7-2 | How Many Ways to Buy a Piece of Land | 25 | 8.1 DFS |
| 7-3 | Left-View of Binary Tree | 25 | 9.2 二叉树的遍历 |
| 7-4 | Professional Ability Test | 30 | 10.4 最短路径+10.6 拓扑排序 |
按通过率来看，第一题最难。。。本次考试的第一题以一己之力大大拉高了整体难度。千万别小看第一题，很可能就来个下马威。第二题前缀和可能不容易想到，【A1046 Shortest Distance】这题也用到了该思想。即使想不出来，万能的DFS也很好用。第三、四题都是套路题，只是小小变通一下。尽管如此，因为第一题的存在，我愿称本次考试为当年最难。