HDU2386Dart Challenge题解动态规划DP

Dart Challenge

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 268    Accepted Submission(s): 134

Problem Description

Clark and Harry are siblings. As they had been rivals since their early childhood, their father decided that both should concentrate on a different sport when they were thirteen. That way, they would not have to compete for success. Now both are twenty years old and excel in different fields: Clark plays chess while Harry participates in dart-tournaments.
Having won a series of three tournaments in a row, Harry started teasing Clark about not having as much success. Clark retorted that chess was less luck-based and thus more difficult. That offended Harry and led him to the reply that in order to play darts optimally, a lot of combinatorics are necessary. Clark returned an icy smile and the comment that memorizing all different late-games could hardly be called “combinatorics”.
This is how it came to the wager. Harry bets that he can find all possible late-games for generalized dart-boards where memorized late-games do not help him. When Clark showed him a list of possible dartboards, Harry had to admit that he probably bit off more than he can chew. As his friend, you have to help him!

A dart-board consists of different areas. Each area has an assigned score for hitting it. Each area also has a double- and a triple-field that are worth twice and three times the score of the area. The only exception is the area for the highest score: It has only a double- and no triple-field! Given the values of the different areas you have to find the number of possible scores that can be obtained with a given number of darts.

 

Input

The inputs start with a line containing a single integer n. Each of the n following lines contains one test case. Each test case starts with two integers 1 <= a <= 100; 1 <= k <= 50, the number of different areas on the
dart-board and the number of darts. a integers 1 <= s_i <= 100 follow. si is the score for hitting area i. All scores are distinct. Remember that each area has a double- and, with exception of the area with the highest score, a triple-field. It is always possible to score 0 with any given dart by not hitting the board.

 

Output

The output for every test case begins with a line containing “Scenario #i:”, where i is the number of the scenario counting from 1. After that, output a single line containing the number of different scores that can be obtained with k darts on the given board. Terminate each test case with an empty line.

 

Sample Input

321 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 252 2 20 101 50 1

 

Sample Output

Scenario #1:172Scenario #2:9Scenario #3:101

 

Source

HDU 2008-10 Public Contest

 

Recommend

lcy

二维01背包计数

我刚开始用一维状态，还把分数拆成三个

背包体积也估计过大，导致爆空间

状态：

d[i][j]表示i个数能否组成j

状态转移方程：

if(d[j-1][k])
     {
      d[j][k]=1;
      d[j][k+a[i]]=1;
      d[j][k+2*a[i]]=1;
      if(p!=i)
       d[j][k+3*a[i]]=1;   
     }

边界：

d[0][0]=1;

代码：

#include<cstdio>#include<cstring>int d[105][20005];int main(){int t,c=1;scanf("%d",&t);while(t--){memset(d,0,sizeof(d));int n,ans=0,m,i,j,k,p=0,x=0,a[105];scanf("%d%d",&n,&m);for(i=0;i<n;i++){scanf("%d",a+i);if(a[i]>a[p])p=i;}x=3*a[p]*m;d[0][0]=1;for(i=0;i<n;i++)for(j=1;j<=m;j++)for(k=0;k<=x;k++)if(d[j-1][k]){d[j][k]=1;d[j][k+a[i]]=1;d[j][k+2*a[i]]=1;if(p!=i)d[j][k+3*a[i]]=1;}for(i=0;i<=x;i++)if(d[m][i])ans++;printf("Scenario #%d:/n",c++);printf("%d/n/n",ans);}}