HDU_01背包系列

HDU_2602 Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 49986    Accepted Submission(s): 20965

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

 

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

分析：裸01背包。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

代码清单：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1000 + 5;int T;int N, V;int value[maxn];int volume[maxn];int dp[maxn];void input() {scanf("%d%d", &N, &V);for(int i = 0; i < N; ++i) {cin >> value[i];}for(int i = 0; i < N; ++i) {cin >> volume[i];}}void solve() {memset(dp, 0, sizeof(dp));for(int i = 0; i < N; ++i) {for(int j = V; j >= volume[i]; --j) {dp[j] = max(dp[j - volume[i]] + value[i], dp[j]);}}cout << dp[V] << endl;}int main() {cin >> T;for(int t = 0; t < T; ++t) {input();solve();}return 0;}

HDU_2546 饭卡

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 22098    Accepted Submission(s): 7730

Problem Description

电子科大本部食堂的饭卡有一种很诡异的设计，即在购买之前判断余额。如果购买一个商品之前，卡上的剩余金额大于或等于5元，就一定可以购买成功（即使购买后卡上余额为负），否则无法购买（即使金额足够）。所以大家都希望尽量使卡上的余额最少。
某天，食堂中有n种菜出售，每种菜可购买一次。已知每种菜的价格以及卡上的余额，问最少可使卡上的余额为多少。

 

Input

多组数据。对于每组数据：
第一行为正整数n，表示菜的数量。n<=1000。
第二行包括n个正整数，表示每种菜的价格。价格不超过50。
第三行包括一个正整数m，表示卡上的余额。m<=1000。

n=0表示数据结束。

 

Output

对于每组输入,输出一行,包含一个整数，表示卡上可能的最小余额。

 

Sample Input

1505101 2 3 2 1 1 2 3 2 1500

 

Sample Output

-4532

 

分析：01背包。涉及贪心策略。首先拿出5元买最贵的菜，那么接下来的背包容量为m-5，物品数量n-1，进行01背包就是了。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2546

代码清单：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1000 + 5;int n, m;int value[maxn];int dp[maxn];void input() {for(int i = 0; i < n; ++i) {cin >> value[i];}cin >> m;}void solve() {if(m < 5) {cout << m << endl;return ;}m -= 5;sort(value, value + n);memset(dp, 0, sizeof(dp));for(int i = 0; i < n - 1; ++i) {for(int j = m; j >= value[i]; --j) {dp[j] = max(dp[j - value[i]] + value[i], dp[j]);}}cout << m + 5 - dp[m] - value[n - 1] << endl;}int main() {while(cin >> n && n != 0) {input();solve();}return 0;}

HDU_2955 Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 20096    Accepted Submission(s): 7443

Problem Description

The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.


For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.


His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.

 

Input

The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj . 
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .

 

Output

For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.

Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

 

Sample Input

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

 

Sample Output

246

 

题意：抢匪抢劫银行j的金额是Mj，被抓的概率是Pj。当被抓概率小于P时，认为抢匪是可以逃脱的。那么问题来了，在可逃脱的情况下，抢匪最多能抢多少钱。

分析：01背包。一眼看去是裸01背包，但是数组下标却是浮点数，所以得想办法转化。我们把银行的金额总和当做背包容量，不被抓的概率当做价值，那么转移方程就是：dp[j] = max(dp[j - p[i]] * (1 - m[i]), dp[j])。那么结果也就显而易见了，找到最大的容量V，dp[V] > 1 - P，那么这个V就是结果了。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2955

代码清单：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define EPS 0.000000001const int maxn = 100 + 5;int T;double P;int N;int p[maxn];double m[maxn];double dp[maxn * maxn];void input() {cin >> P >> N;for(int i = 0; i < N; ++i) {cin >> p[i] >> m[i];}}void solve() {if(P - 0 < EPS) {cout << "0" << endl;return ;}int V = 0;for(int i = 0; i < N; ++i) {V += p[i];}dp[0] = 1;for(int i = 1; i <= V; ++i) dp[i] = 0;for(int i = 0; i < N; ++i) {for(int j = V; j >= p[i]; --j) {dp[j] = max(dp[j - p[i]] * (1 - m[i]), dp[j]);}}for(int i = V; i >= 0; --i) { if(dp[i] - (1 - P) > EPS) {cout << i << endl;return ;}}}int main() {cin >> T;for(int t = 0; t < T; ++t) {input();solve();}return 0;}

 

HDU_1203 I NEED A OFFER!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 24189    Accepted Submission(s): 9728

Problem Description

Speakless很早就想出国，现在他已经考完了所有需要的考试，准备了所有要准备的材料，于是，便需要去申请学校了。要申请国外的任何大学，你都要交纳一定的申请费用，这可是很惊人的。Speakless没有多少钱，总共只攒了n万美元。他将在m个学校中选择若干的（当然要在他的经济承受范围内）。每个学校都有不同的申请费用a（万美元），并且Speakless估计了他得到这个学校offer的可能性b。不同学校之间是否得到offer不会互相影响。“I NEED A OFFER”，他大叫一声。帮帮这个可怜的人吧，帮助他计算一下，他可以收到至少一份offer的最大概率。（如果Speakless选择了多个学校，得到任意一个学校的offer都可以）。

 

Input

输入有若干组数据，每组数据的第一行有两个正整数n,m(0<=n<=10000,0<=m<=10000) 
后面的m行，每行都有两个数据ai(整型),bi(实型)分别表示第i个学校的申请费用和可能拿到offer的概率。 
输入的最后有两个0。

 

Output

每组数据都对应一个输出，表示Speakless可能得到至少一份offer的最大概率。用百分数表示，精确到小数点后一位。

 

Sample Input

10 34 0.14 0.25 0.30 0

 

Sample Output

44.0%
Hint
You should use printf("%%") to print a '%'.
 

 

分析：01背包。此题跟HDU_2955很像，求可能得到至少一份工作的最大概率，那么可以考虑反面，得不到工作的最小概率。那么就很容易理解了，转移方程：dp[j] = min(dp[j - a[i]] * (1 - b[i]), dp[j])。结果自然而然就是 1 - dp[n]了。注意这里输出百分数，记得转化。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1203

代码清单：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 10000 + 5;int n, m;int a[maxn];double b[maxn];double dp[maxn];void input() {for(int i = 0; i <m; ++i) {scanf("%d%lf", &a[i], &b[i]);}}void solve() {dp[0] = 1;for(int i = 1; i <= n; ++i) dp[i] = 1;for(int i = 0; i < m; ++i) {for(int j = n; j >= a[i]; --j) {dp[j] = min(dp[j - a[i]] * (1 - b[i]), dp[j]);//cout << dp[j] << " ";}//cout << endl;}printf("%.1lf%%\n", (1 - dp[n]) * 100);}int main() {while(scanf("%d%d", &n, &m) != EOF && n+m) {input();solve();}return 0;}

HDU_1171 Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 35145    Accepted Submission(s): 12188

Problem Description

Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

 

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

 

Output

For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

 

Sample Input

210 120 1310 1 20 230 1-1

 

Sample Output

20 1040 40

 

题意：把所有电脑分成A，B部分，使得A、B两部分金额尽可能相近，并保证金额 A > B。

分析：01背包。把总价平分一下，较小的一部分当做B背包的容量，那么对B进行01背包即可得到B的最大价值，A的价值也就出来了。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1171

代码清单：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1000 + 5;const int maxm = 100 + 5;int N;int V[maxn];int M[maxn];int value[maxn * maxm];int dp[maxn * maxm];void input() {for(int i = 0; i < N; ++i) {scanf("%d%d", &V[i], &M[i]);}}void solve() {int sumV = 0, va = 0;for(int i = 0; i < N; ++i) {sumV += V[i] * M[i];for(int j = 0; j < M[i]; ++j) {value[va++] = V[i];}}int minV = sumV / 2;memset(dp, 0, sizeof(dp));for(int i = 0; i < va; ++i) {for(int j = minV; j >= value[i]; --j) {dp[j] = max(dp[j - value[i]] + value[i], dp[j]);}}printf("%d %d\n", sumV - dp[minV], dp[minV]);}int main() {while(scanf("%d", &N) != EOF && N >= 0) {input();solve();}return 0;}

HDU_2639 Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3967    Accepted Submission(s): 2051

Problem Description

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link:

Here is the link:http://acm.hdu.edu.cn/showproblem.php?pid=2602

Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum.

If the total number of different values is less than K,just ouput 0.

 

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

Output

One integer per line representing the K-th maximum of the total value (this number will be less than 2³¹).

 

Sample Input

35 10 21 2 3 4 55 4 3 2 15 10 121 2 3 4 55 4 3 2 15 10 161 2 3 4 55 4 3 2 1

 

Sample Output

1220

 

题意：与HDU_2602也就是本博文第一题一样，只不过这次是求第K大优解。

分析：01背包第k优解。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2639

代码清单：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 100 + 5;const int maxk = 30 + 5;const int maxv = 1000 + 5;int T;int N, V, K;int value[maxn];int volume[maxn];int dp[maxv][maxk];int pre[maxk], now[maxk];void input() {cin >> N >> V >> K;for(int i = 0; i < N; ++i) {cin >> value[i];}for(int i = 0; i < N; ++i) {cin >> volume[i];}}void solve() {memset(dp, 0, sizeof(dp));pre[K + 1] = now[K + 1] = -1;for(int i = 0; i < N; ++i) {for(int j = V; j >= volume[i]; --j) {for(int k = 1; k <= K; ++k) {pre[k] = dp[j][k];now[k] = dp[j - volume[i]][k] + value[i];}int idxp = 1, idxn = 1, idxk = 1;while(idxk <= K && (idxp <= K || idxn <= K)) {if(pre[idxp] >= now[idxn]) {dp[j][idxk] = pre[idxp++];}else {dp[j][idxk] = now[idxn++];}//Remove duplicate numberif(idxk == 1 || dp[j][idxk] != dp[j][idxk - 1]) { idxk += 1;}}}}cout << dp[V][K] << endl;}int main() {cin >> T;for(int t = 0; t < T; ++t) {input();solve();}return 0;}