# 题目2 : Tree Restoration

## 描述

There is a tree of N nodes which are numbered from 1 to N. Unfortunately, its edges are missing so we don’t know how the nodes are connected. Instead we know:

1. Which nodes are leaves
2. The distance (number of edges) between any pair of leaves
3. The depth of every node (the root’s depth is 1)
4. For the nodes on the same level, their order from left to right

Can you restore the tree’s edges with these information? Note if node u is on the left of node v, u’s parent should not be on the right of v’s parent.

## 输入

The first line contains three integers N, M and K. N is the number of nodes. M is the depth of the tree. K is the number of leaves.

The second line contains M integers A1, A2, … AM. Ai represents the number of nodes of depth i.

Then M lines follow. The ith of the M lines contains Ai numbers which are the nodes of depth i from left to right.

The (M+3)-th line contains K numbers L1, L2, … LK, indicating the leaves.

Then a K × K matrix D follows. Dij represents the distance between Li and Lj.

1 ≤ N ≤ 100

## 输出

For every node from 1 to N output its parent. Output 0 for the root’s parent.

8 3 5
1 3 4
1
2 3 4
5 6 7 8
3 5 6 7 8
0 3 3 3 3
3 0 2 4 4
3 2 0 4 4
3 4 4 0 2
3 4 4 2 0

0 1 1 1 2 2 4 4

# 解析

• 对于每层第一个节点，上一层的第一个非叶子节点必然是这个节点的父节点。
• 此时记录该节点的父节点，并把父节点的子节点记录为当前节点，并以此更新这个父节点与其他已知节点的距离。
• 然后看该层下一个节点。
• 下一个节点若与前一个节点距离为2，则两者父节点相同。
• 若大于2，则该节点的父节点是上一层的下一个非叶子节点。同时记录父、子节点，更新距离。
• 如此每层循环，每层结束进入上一层，直到第二层。（第一层根节点不必再判断了。

## AC代码（C++）：

#include<iostream>
using namespace std;

int main() {
int N, M, K;
//N is the number of nodes. M is the depth of the tree. K is the number of leaves.
cin >> N >> M >> K;

int depth[105] = { 0 };
//depth[] represents the number of nodes of depth[i].
for (int i = 0; i < M; i++) {
cin >> depth[i];
}

int tree[105][105] = { 0 };
//the nodes of depth[i] from left to right.
for (int i = 0; i < M; i++) {
for (int j = 0; j < depth[i]; j++) {
cin >> tree[i][j];
}
}

int leaves[105] = { 0 };
//leaves node
int is_leaves[105] = { 0 };
//Determine whether a node is a leaf node
for (int i = 0; i < K; i++) {
cin >> leaves[i];
is_leaves[leaves[i]] = 1;
}

int dis[105][105] = { 0 };
// the distance between Li and Lj.
for (int i = 0; i < K; i++) {
for (int j = 0; j < K; j++) {
cin >> dis[leaves[i]][leaves[j]];
}
}

int father[105] = { 0 };
int son[105] = { 0 };
//record a child node of a node

father[tree[0][0]] = 0;
for (int i = 0; i < depth[1]; i++) {
father[tree[1][i]] = tree[0][0];
}

for (int i = M - 1; i >= 2; i--) {
int up_point = 0;
////i-1层中第一个不是叶子节点的节点，必定是当前节点的父亲
while (is_leaves[tree[i - 1][up_point]] && up_point < depth[i - 1]) {
up_point++;
}
int point_a = tree[i][0];
father[point_a] = tree[i - 1][up_point];
son[tree[i - 1][up_point]] = point_a;
//更新距离
for (int k = 0; k < N; k++) {
if (dis[point_a][k] > 0) {
dis[father[point_a]][k] = dis[k][father[point_a]] = dis[point_a][k] - 1;
}
}

for (int k = 0; k < K; k++) {
dis[leaves[k]][father[point_a]] = dis[father[point_a]][leaves[k]] = dis[leaves[k]][point_a] - 1;
}

for (int j = 1; j < depth[i]; j++) {
int point_b = tree[i][j];
//如果和前一个节点u距离为2，说明父亲节点相同
if (dis[point_a][point_b] == 2) {
father[point_b] = father[point_a];
}
//否则，父亲节点是i-1层中下一个非叶子节点
else {
up_point++;
while (is_leaves[tree[i - 1][up_point]] && up_point < depth[i - 1]) {
up_point++;
}
father[point_b] = tree[i - 1][up_point];
son[tree[i - 1][up_point]] = point_b;
//更新距离
for (int kk = 0; kk < N; kk++) {
if (dis[point_b][kk] > 0) {
dis[father[point_b]][kk] = dis[kk][father[point_b]] = dis[point_b][kk] - 1;
}
}
}
point_a = point_b;
}
//更新i-1层相邻节点的距离
for (int kk = 0; kk < depth[i - 1] - 1; kk++) {
if (!is_leaves[tree[i - 1][kk]] && !is_leaves[tree[i - 1][kk + 1]]) {
dis[tree[i - 1][kk]][tree[i - 1][kk + 1]] = dis[tree[i - 1][kk + 1]][tree[i - 1][kk]] = dis[son[tree[i - 1][kk]]][son[tree[i - 1][kk + 1]]] - 2;
}
}
}
cout << father[1];
for (int i = 2; i <= N; i++) {
cout << " " << father[i];
}
return 0;
}


## WA代码（C++）：

100个节点的数据太多了。暂时无法测试。

#include<iostream>
using namespace std;

int main() {
int N, M, K;
//N is the number of nodes. M is the depth of the tree. K is the number of leaves.
cin >> N >> M >> K;

int count_depth[105] = { 0 };
for (int i = 1; i <= M; i++) {
cin >> count_depth[i];
}

int tree[105][105] = { 0 };
int depth_of_node[105] = { 0 };
for (int i = 1; i <= M; i++) {
for (int j = 1; j <= count_depth[i]; j++) {
cin >> tree[i][j];
depth_of_node[tree[i][j]] = i;
}
}

int is_leaf[105] = { 0 };
int num_leaf[105] = { 0 };
for (int i = 1; i <= K; i++) {
int TEMP;
cin >> TEMP;
num_leaf[i] = TEMP;
is_leaf[TEMP] = 1;
}

int dis[105][105] = { 0 };
for (int i = 1; i <= K; i++) {
for (int j = 1; j <= K; j++) {
cin >> dis[num_leaf[i]][num_leaf[j]];
}
}

int father_node[105] = { 0 };

int s_son_node[105] = { 0 };
for (int i = 1; i <= N; i++) {
if (is_leaf[i]) {
s_son_node[i] = i;
}
}

for (int i = M; i > 0; i--) {
int count_top = 1;
for (int j = 1; j <= count_depth[i]; j++) {
if (j == 1) {
for (; count_top <= count_depth[i - 1]; count_top++) {
if (is_leaf[tree[i - 1][count_top]] != 1) {
father_node[tree[i][j]] = tree[i - 1][count_top];
s_son_node[tree[i - 1][count_top]] = tree[i][j];
count_top++;
break;
}
}
continue;
}
if (dis[tree[i][j]][tree[i][j - 1]] == 2) {
father_node[tree[i][j]] = father_node[tree[i][j - 1]];
}
else {
for (; count_top <= count_depth[i - 1]; count_top++) {
if (is_leaf[tree[i - 1][count_top]] != 1) {
father_node[tree[i][j]] = tree[i - 1][count_top];
s_son_node[tree[i - 1][count_top]] = tree[i][j];
count_top++;
break;
}
}
continue;
}
}

//更新上一层节点间距离
for (int j = 1; j <= count_depth[i - 1]; )
{
for (int jj = j + 1; jj <= count_depth[i - 1]; jj++)
{
dis[tree[i - 1][j]][tree[i - 1][jj]] = dis[tree[i - 1][jj]][tree[i - 1][j]] = dis[s_son_node[tree[i - 1][j]]][s_son_node[tree[i - 1][jj]]] - (depth_of_node[s_son_node[tree[i - 1][j]]] - depth_of_node[tree[i - 1][j]]) - (depth_of_node[s_son_node[tree[i - 1][jj]]] - depth_of_node[tree[i - 1][jj]]);
}
j++;
}
}
//输出
for (int i = 1; i < N; i++)
{
cout << father_node[i] << " ";
}
cout << father_node[N];

return 0;
}


## 最终结果

1490Tree RestorationACG++5ms0MB

# 题目1 : Legendary Items

## 描述

Little Hi is playing a video game. Each time he accomplishes a quest in the game, Little Hi has a chance to get a legendary item.

At the beginning the probability is P%. Each time Little Hi accomplishes a quest without getting a legendary item, the probability will go up Q%. Since the probability is getting higher he will get a legendary item eventually.

After getting a legendary item the probability will be reset to ⌊P/(2I)⌋% (⌊x⌋ represents the largest integer no more than x) where I is the number of legendary items he already has. The probability will also go up Q% each time Little Hi accomplishes a quest until he gets another legendary item.

Now Little Hi wants to know the expected number of quests he has to accomplish to get N legendary items.

Assume P = 50, Q = 75 and N = 2, as the below figure shows the expected number of quests is 3.25

2*50%*25% + 3*50%*75%*100% + 3*50%*100%*25% + 4*50%*100%*75%*100% = 3.25


## 输入

The first line contains three integers P, Q and N.

1 \leq N \leq 10^6, 0 \leq P \leq 100, 1 \leq Q \leq 1001≤N≤106,0≤P≤100,1≤Q≤100

## 输出

Output the expected number of quests rounded to 2 decimal places.

50 75 2

3.25

# 解析

EX=\sum{(X \cdot P(X))}EX=∑(X⋅P(X))

E(X+Y)=EX+EYE(X+Y)=EX+EY

1.初始任务数（至少1次任务就能获得）numQuests=1numQuests=1
2.第一次任务对该次获得传奇物品增加的期望：incE=(1-P)incE=(1−P)
3.调整概率P=P+QP=P+Q
4.进行下一次任务对该次获得传奇物品增加的期望计算：incE=incE\cdot(1-P)incE=incE⋅(1−P)
5.任务期望：numQuests=numQuest+incEnumQuests=numQuest+incE
6.回到步骤3
7.直至P=P+Q>100P=P+Q>100终止此次获得传奇物品的期望计算numQuestsnumQuests

## AC代码（C++）：

#include<iostream>
#include<iomanip>

using namespace std;

int main()
{
int P, Q, N;
cin >> P >> Q >> N;

double count_q=Q*1.00/100;
double count_p;
double result = 0.00;

for (int i = 1; i <= N; i++)
{
double next_result = 1;

count_p = P*1.00 / 100;
double E1 = 1.00;
while (1)
{
E1 = E1*(1.00 - count_p);
next_result += E1;
count_p += count_q;
if (count_p > 1.00)
break;
}

result += next_result;
P = P / 2;
}

cout << fixed << setprecision(2) << result;

return 0;
}


## 最终结果

1489Legendary ItemsACG++22ms0MB