7-1 Panda and PP Milk (20分)
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 (≤10^4) is given as the number of pandas. Then in the next line, npositive 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 140Sample Output:
3000Hint:
The distribution of milk is the following:
400 300 200 500 400 300 200 200 200 300参考代码：
#include
#define ll long long
using namespace std;
const int maxn=1e5+5;
int n;
int a[maxn],val[maxn];
int main(){
scanf("%d",&n);
for(int i=1; i<=n; ++i) scanf("%d",&a[i]);
int cur=10000;
for(int i=1; i<=n; ++i){
if(a[i]>cur) val[i]=val[i-1]+100;
else if(a[i]==cur) val[i]=val[i-1];
else val[i]=200;
cur=a[i];
}
cur=10000;
for(int i=n; i>=1; --i){
if(a[i]>cur) val[i]=max(val[i+1]+100,val[i]);
else if(a[i]==cur) val[i]=max(val[i],val[i+1]);
else val[i]=max(val[i],200);
cout << val[i] << endl;
cur=a[i];
}
int ans=0;
for(int i=1; i<=n; ++i) ans+=val[i];
printf("%d\n",ans);
return 0;
} 7-2 How Many Ways to Buy a Piece of Land (25分)
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 (≤10^4), the number of pieces of the land (hence the land pieces are numbered from 1 to N in order), and M (≤10^9), 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 10^9.
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 9Sample Output:
11Hint:
The 11 different ways are:
38
42
15
24
9
38 42
42 15
42 15 24
15 24
15 24 9
24 9参考代码：
#include
#define ll long long
using namespace std;
const int maxn=1e5+5;
int n,m;
int a[maxn],pref[maxn];
int main(){
scanf("%d%d",&n,&m);
for(int i=1; i<=n; ++i) scanf("%d",&a[i]),pref[i]=pref[i-1]+a[i];
int ans=0;
for(int i=1; i<=n; ++i){
int L=i,R=n;
while(L>1;
if(pref[mid]-pref[i-1]<=m) L=mid;
else R=mid-1;
}
if(pref[L]-pref[i-1]<=m) ans+=L-i+1;
}
printf("%d\n",ans);
return 0;
} 7-3 Left-View of Binary Tree (25分)
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 8Sample Output:
1 2 3 4 5参考代码：
#include
#include
using namespace std;
struct node{
int data, level;
node *left, *right; //此处命名注意*和后面的俩连在一起
};
int n;
int in[25], pre[25];
bool level_print[25] = {false};
node* build(int inL, int inR, int preL, int preR){
if(inL > inR) return NULL;
node* root = new node;
int pos = inL;
for(;pos <= inR; pos++){
if(in[pos] == pre[preL]) break;
}
root->data = pre[preL];
root->left = build(inL, pos-1, preL+1, preL+1+(pos-1-inL));
root->right = build(pos+1, inR, preR-(inR-pos-1), preR);
return root;
}
void BFS(node* root){
queue q;
root->level = 1;
q.push(root);
printf("%d", root->data);
level_print[root->level] = true;
while(!q.empty()){
node* top = q.front();
q.pop();
if(!level_print[top->level]){
printf(" %d", top->data);
level_print[top->level] = true;
}
if(top->left != NULL){
top->left->level = top->level+1;
q.push(top->left);
}
if(top->right != NULL){
top->right->level = top->level+1;
q.push(top->right);
}
}
}
int main(){
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &in[i]);
for(int i = 0; i < n; i++) scanf("%d", &pre[i]);
node* root = new node;
root = build(0, n-1, 0, n-1);
BFS(root);
return 0;
}
7-4 Professional Ability Test (30分)
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 Dwhere 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->...->THowever, 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 7Sample 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->7Sample 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 1Sample Output 2:
Impossible.
You may take test 3 directly.
Error.思路分析：
这种30分的题目，读题并理解清楚题意真的很重要。模拟训练时就因为没有真正读懂导致无法下手。完成本题主要从两方面入手：1.判断给定结点是否在环中。2.寻找入度为零的结点到达每一给定结点的最短路径。（这里刚开始理解错了，不是连着判断，不是完成后一个结点必须先完成前面的结点，而是每一给定结点单独判断寻找最短路径）
一些易错点：给定的通过考试的分数以及所给奖励均由两个点来限定，实际意义均是边权，不要想当然地将其中某一个视为点权；访问某一点不要求入度为零，但某点在环中则不能访问。个人感觉这有点小问题，因为既然在环中不能被访问，那么应该有入度为零才能被访问的要求，否则该环可以被打破（即通过不在环中的前驱关系来打破）。但还是要根据题意来。
判环
方法比较多，介绍三种：
1.对每一给定结点进行深搜，并将该结点作为形参穿入每次递归，若在深搜过程中再次遇到该结点，即该结点在环中，不过这种方法对每个结点都要进行一次深搜，比较耗时。
2.对整个图进行处理，寻找入度为零的点，删去并对其连接的另一个结点的入度减一，不断重复这一过程，那么最后入度不为零的点即都在环中。（一般建议以队列实现，当然遍历每个结点来处理也可以）
寻找最短路径
此次的寻找最短路径与以往有所区别，这里是给定目标结点，寻找一个源点使其路径最短。也即多源点，单目标结点。
几种方向：
1.逆向思维即可，将该目标结点视为源点进行算法，进行完后再依次遍历每一个入度为零的结点，找到符合题意的最短路径。
2.人为构造一个超级源点，指向所有入度为零的结点，且边权均设置为0。这样多源点问题就转换为单源点问题。
3.使用SPFA算法，构建队列，将所有入度为零的点入队，且每出队时都看是否有新的入度为零的结点产生，产生则入队。这样，既可以判断是否有环，同时也解决了多源问题。
参考代码：
#include
#include
#include
#include
using namespace std;
const int maxv = 1010;
const int INF = 0x7fffffff;
int n, m, k;
int in[maxv] = {0}, indegree[maxv] = {0};
int w[maxv], c[maxv], seq[maxv], pre[maxv];
int G[maxv][maxv], money[maxv][maxv];
vector post[maxv]; //记录后继结点
void SPFA(){
queue q;
fill(pre, pre+maxv, -1);
fill(c, c+maxv, INF);
fill(w, w+maxv, 0);
for(int i = 0; i < n; i++){
if(!indegree[i]){//源点初始化并入队
c[i] = 0;
w[i] = 0;
q.push(i);
}
}
while(!q.empty()){
int u = q.front();
q.pop();
for(int j = 0; j < post[u].size(); j++){
int v = post[u][j];
indegree[v]--;
if(!indegree[v]) q.push(v);
if(c[u] + G[u][v] < c[v]){
c[v] = c[u] + G[u][v];
w[v] = w[u] + money[u][v]; //此处对权值的更新不要忘
pre[v] = u;
}else if(c[u] + G[u][v] == c[v]){
if(w[u] + money[u][v] > w[v]){
w[v] = w[u] + money[u][v]; //注意不要忘了这句话，即更新最大权值（幸福值）
pre[v] = u;
}
}
}
}
}
int main(){
int T1, T2, S, D;
int flag = 1;
scanf("%d %d", &n, &m);
fill(G[0], G[0]+maxv*maxv, INF);
for(int i = 0; i < m; i++){
scanf("%d %d %d %d", &T1, &T2, &S, &D);
indegree[T2]++;
in[T2]++;
G[T1][T2] = S;
money[T1][T2] = D;
post[T1].push_back(T2);
}
scanf("%d", &k);
for(int i = 0; i < k; i++) scanf("%d", &seq[i]);
SPFA();
for(int i = 0; i < k; i++){
if(indegree[seq[i]]){
flag = 0;
break;
}
}
if(flag){
printf("Okay.\n");
for(int i = 0; i < k; i++){
vector ans;
for(int pos = seq[i]; pos != -1; pos = pre[pos]){
ans.push_back(pos);
}
reverse(ans.begin(), ans.end());
if(!in[seq[i]]) printf("You may take test %d directly.\n", seq[i]);
else for(int j = 0; j < ans.size(); j++) printf("%d%s", ans[j], j == ans.size()-1 ? "\n":"->");
}
}else{
printf("Impossible.\n");
for(int i = 0; i < k; i++){
if(!in[seq[i]]) printf("You may take test %d directly.\n", seq[i]);
else printf("Error.\n");
}
}
return 0;
} 本次考试几点收获：
1.英语阅读理解能力真的很重要，题意理解不清楚，代码怎么写都不对。
2.#include
3.存储图时，使用邻接表要比使用邻接矩阵节省很多内存，但使用邻接表时记得要将边的相应信息存到一个结构体里面去。
4.通过记录前驱来生成最终输出路径时，可以使用深搜来逆序输出，当然使用reverse也很简洁。
5.SPFA算法比Dijkstra更简洁，使用范围也更广，及其推荐使用。