Strategic game
Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a 阅读全文 »
ural1018
22
9月
2015
1018. Binary Apple Tree
Let's imagine how apple tree looks in binary computer world. You're right, it looks just like a binary tree, i.e. any biparous branch splits up to exactly two new branches. We will enumerate by integers the root of binary apple tree, points of branching and the ends of twigs. This way we may distinguish different branches by their ending points. We will assume that root of tree always is numbered by 1 and all numbers used for enumerating are numbered in range from 1 to N, where N is the total number of all enumerated points. For instance in the picture below N is equal to 5. Here is an example of an enumerated tree with four branches:
InputAs you may know it's not convenient to pick an apples from a tree when there are too much of branches. That's why some of them should be removed from a tree. But you are interested in removing branches in the way of minimal loss of apples. So your are given amounts of apples on a branches and amount of branches that should be preserved. Your task is to determine how many apples can remain on a tree after removing of excessive branches.
First line of input contains two numbers: N and Q (2 ≤ N ≤ 100; 1 ≤ Q ≤ N − 1). N denotes the number of enumerated points in a tree. Q denotes amount of branches that should be preserved. NextN − 1 lines contains descriptions of branches. Each description consists of a three integer numbers divided by spaces. The first two of them define branch by it's ending points. The third number defines the number of apples on this branch. You may assume that no branch contains more than 30000 apples.
Output
Output should contain the only number — amount of apples that can be preserved. And don't forget to preserve tree's root 😉
Sample
| input | output |
| 5 21 3 11 4 10
2 3 20 3 5 20 |
21 |
题目类型：树形DP
算法分析：dp[i][j]表示以i为根的子树恰好保留j个分支所具有的最多苹果数量，初始化dp全为0。最后直接输出dp[1][m]即可，其中m为保留的分支的个数。状态转移方程为：dp[u][i] = max (dp[u][i], dp[u][i-k] + dp[v][k-1] + w)，其中v为u的子节点，w为u-v这条边上的苹果数
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#pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 100 + 66; int dp[maxn][maxn]; bool vis[maxn]; int n, m; //分别表示顶点v。w和下一个邻接的位置 struct Node { int v, w, next; }; Node edge[maxn*maxn]; int head[maxn], edgelen;//分别表示顶点的表头节点和边的数量 //将所有的数组均初始化为-1 void Init () { memset (head, -1, sizeof (head)); memset (edge, -1, sizeof (edge)); edgelen = 0; } //使用头插法构建 void AddEdge (int u, int v, int w) { edge[edgelen].v = v; edge[edgelen].w = w; edge[edgelen].next = head[u]; head[u] = edgelen++; } void dfs (int rt) { vis[rt] = true; for (int i = head[rt]; i != -1; i = edge[i].next) { if (vis[edge[i].v]) continue; dfs (edge[i].v); for (int j = m; j >= 1; j--) for (int k = 1; k <= j; k++) dp[rt][j] = max (dp[rt][j], dp[rt][j-k] + dp[edge[i].v][k-1] + edge[i].w); } } int main() { // CPPFF; while (cin >> n >> m) { int u, v, w; Init(); for (int i = 1; i <= n - 1; i++) { cin >> u >> v >> w; AddEdge (u, v, w); AddEdge (v, u, w); } memset (vis, false, sizeof (vis)); memset (dp, 0, sizeof (dp)); dfs (1); cout << dp[1][m] << endl; } return 0; } |
poj2063
22
9月
2015
Investment
John never knew he had a grand-uncle, until he received the notary's letter. He learned that his late grand-uncle had gathered a lot of money, somewhere in South-America, and that John was the only inheritor.
John did not need that much money for the moment. But he realized that it would be a good idea to store this capital in a safe place, and have it grow until he decided to retire. The bank convinced him that a certain kind of bond was interesting for him.
This kind of bond has a fixed value, and gives a fixed amount of yearly interest, payed to the owner at the end of each year. The bond has no fixed term. Bonds are available in different sizes. The larger ones usually give a better interest. Soon John realized that the optimal set of bonds to buy was not trivial to figure out. Moreover, after a few years his capital would have grown, and the schedule had to be re-evaluated.
Assume the following bonds are available:
| Value |
Annual |
| 4000 3000 |
400 250 |
With a capital of e10 000 one could buy two bonds of $4 000, giving a yearly interest of $800. Buying two bonds of $3 000, and one of $4 000 is a better idea, as it gives a yearly interest of $900. After two years the capital has grown to $11 800, and it makes sense to sell a $3 000 one and buy a $4 000 one, so the annual interest grows to $1 050. This is where this story grows unlikely: the bank does not charge for buying and selling bonds. Next year the total sum is $12 850, which allows for three times $4 000, giving a yearly interest of $1 200.
Here is your problem: given an amount to begin with, a number of years, and a set of bonds with their values and interests, find out how big the amount may grow in the given period, using the best schedule for buying and selling bonds.
Input
The first line contains a single positive integer N which is the number of test cases. The test cases follow.
The first line of a test case contains two positive integers: the amount to start with (at most $1 000 000), and the number of years the capital may grow (at most 40).
The following line contains a single number: the number d (1 <= d <= 10) of available bonds.
The next d lines each contain the description of a bond. The description of a bond consists of two positive integers: the value of the bond, and the yearly interest for that bond. The value of a bond is always a multiple of $1 000. The interest of a bond is never more than 10% of its value.
Output
For each test case, output – on a separate line – the capital at the end of the period, after an optimal schedule of buying and selling.
Sample Input
1
10000 4
2
4000 400
3000 250
Sample Output
14050
Source
题目类型：完全背包
算法分析：直接使用d次完全背包求解即可，注意此时题目已经给出所有的股票的值都是1000的倍数，所以此时要将股票的值除上1000，这样才能优化时间复杂度！！！
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#pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define LL long long #define ULL unsigned long long #define DB(ccc) cout << #ccc << " = " << ccc << endl const int INF = 0x7FFFFFFF; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 3e5 + 66; int v[66], w[66], dp[maxn]; int main() { // CFF; int t; scanf ("%d", &t); while (t--) { int V, d, n; scanf ("%d%d%d", &V, &d, &n); for (int i = 1; i <= n; i++) { scanf ("%d%d", &v[i], &w[i]); v[i] /= 1000; } int maxval; for (int i = 1; i <= d; i++) { memset (dp, 0, sizeof (dp)); for (int j = 1; j <= n; j++) for (int k = v[j]; k <= V / 1000; k++) dp[k] = max (dp[k], dp[k-v[j]] + w[j]); maxval = -INF; for (int i = 0; i <= V / 1000; i++) maxval = max (maxval, dp[i]); V += maxval; } printf ("%d\n", V); } return 0; } |