poj 1013 great equipment（背包，dp）

A - Great Equipment

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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Description

Once upon a time, there lived Catherine Ironfist, the Queen of Enroth. One day, she received the news of her father's death. So she sailed for Erathia to attend her father's funeral. Fearing the worst, she assembled a military fleet as her escort. On reaching the coast of Erathia, Catherine found an allied wizard's tower, devastated from battle and abandoned. There she learned that a black-hearted knight poisoned her father using a goblet of wine, and Erathia was falling to the enemies. And then, she mustered local armies, and marched to Erathia's castle, restoring lost land along the way.

During the battles, she found that the equipments for the soldiers were in urgent need. And she knew clearly that the best equipments were made by the battle dwarf's workshop in the world. The workshop's equipments were well known for the firmness and top-quality. ``Cloak of the Undead King", ``Armor of the Damned", ``Angelic Helm" are the nonesuch ones. But unfortunately, the workshop was seated at the Erathia's castle, the territory under the enemy's control. So she sent a brave volunteer to come into the castle and asked for the workshop's help.

``It's our pleasure to help the righteous heroine." Rion, the leader of the workshop sent the message to Catherine, `` We haven't enough resources to build the nonesuch equipments. So we'll try to offer the ordinary equipments as more as possible. Still, those ones are much better the equipments made by other workshops. But we have faced a difficult problem. The castle is in a state of siege. The guards prohibited the equipments to be carried away from the castle. We have to ask for the trade caravans' help. As you know, each trade caravan's capability of carrying equipments is limited. If they had carried a little more, they would be detected by the guards, which would lead them into the prison. So you have to arrange how to carry these equipments."

The workshop would offer helms, armors and boots. These three ones had different defend capabilities. Also, their weight and size were different from each other. What's more, Rion had told Catherine that if armed with those three ones together, they would form a set of equipments, which might provide much more defend capability. As far as the trade caravan was concerned, each one had its special weight limitation and size limitation. Catherine had to make the decision on how to arrange the transportation plan to provide her soldiers as more defend capabilities as possible. Could you help her to finish the plan?

Input 

The input describes several test cases. The first line of input for each test case contains a single integer n , the number of trade caravans(0n100) .

The following four lines describe the information of those equipments. The first line contains three integers w₁ , s₁ and d₁ , indicating the weight, size and defend capabilities of the helm. The integers w₂ , s₂ and d₂ in the second line represent the weight, size and defend capabilities of the armor. Also, in the third line, w₃ , s₃ and d₃ are the weight, size and defend capabilities of the boot. The fourth line contains four integers c₁ , c₂ , c₃ and d₄ . Among those integers, c₁ , c₂ , c₃ are the number of helms, armors and boots in a set of equipments, d4 is the capability of this set.

In the test case, following those data are n lines, describing the carrying capabilities of the trade caravans. Each line contains two integers,x_i and y_i , indicating the weight limit and size limit of a trade caravan.

The input is terminated by a descri_ption starting with n = 0. This descri_ption should not be processed.

Note: Because of the trade caravans' carrying capabilities, you may assume the quantities of the helms, armors and boots will not exceed 500 respectively.

Output 

Your program must compute the defend capability of the best carrying plan. That is, after having performed the carrying plan, the defend capability of the equipments which have been carried away from the castle should be the largest. For each test case in the input file, print the case number and a colon, and then the defend capability of those equipments.

Print a blank line between test cases.

Sample Input 

31 1 35 6 102 1 21 1 1 501 15 62 10

Sample Output 

Case 1: 50

设f[n][x][y] 表示前n辆车装了x个装备1和y个装备2之后能装的最多的装备3的个数。

递推关系如下：

f[n][x][y] = max ( f[n - 1][x - a][y - b] + c);    其中f[0][x][y]=0

 

其中a和b的范围为第n辆车可以装的装备1和装备2的个数, c 为第n辆车装完a个装备1和b个装备2后可以装的装备3的个数。

通过递推求得f[n][x][y]后，可以遍历x和y（x，y的范围可以在上述递推的过程中得到）求得最终的结果。

由于f[n]只和f[n-1]有关，所以实际上 f 数组的第一维只需要2就可以(详见代码)

显然，需要先得到可能是最优解的x,y,z的各种组合，再通过比较得到的最大总value。在相同x,y的情况下，我们总是尽量使z最大。直接用搜索必然很耗时间，这里我们用DP来求x,y组合对应的z的最大值。状态state[m][x][y]表示在使用1-m这m个背包的情况下，取x,y值时，对应的z的最大值。即有状态方程state[m][x][y]=max(state[m-1][x'][y']+zi),其中(xi+x'=x,yi+y'=y,xi*wx+yi*wy+zi*wz<=w[m],xi*sx+yi*sy+zi*sz<=s[m])（因为要求zi要尽量大，所以同时会有xi*wx+yi*wy+（zi+1）*wz>w[m]，xi*sx+yi*sy+（zi+1）*sz>s[m]）.

最后ans=max(cal_value(x,y,state[n][x][y]))。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define rg(x,y) (x=x>y?x:y)#define min(x,y) (x>y?y:x) int dp[2][501][501],w[501],s[501];int main(){int cs=0,n;int wx,wy,wz,sx,sy,sz,vx,vy,vz,c1,c2,c3,def;while(cin>>n&&n){cin>>wx>>sx>>vx>>wy>>sy>>vy>>wz>>sz>>vz>>c1>>c2>>c3>>def;for(int i=1;i<=n;i++) cin>>w[i]>>s[i];memset(dp[0],0,sizeof(dp[0]));int ix,iy,iz,tempx,tempy,temp,maxx=0,maxy=0;for(int i=1;i<=n;i++){memset(dp[1],0xff,sizeof(dp[1]));tempx=min(w[i]/wx,s[i]/sx); for(ix=0;ix<=tempx;ix++)  {  tempy=min((w[i]-ix*wx)/wy,(s[i]-ix*sx)/sy);  for(iy=0;iy<=tempy;iy++)  {  iz=min((w[i]-ix*wx-iy*wy)/wz,(s[i]-ix*sx-iy*sy)/sz);  for(int a=0;a<=maxx;a++)  {  for(int b=0;b<=maxy;b++)  {  if(dp[0][a][b]>=0)  {  rg(dp[1][a+ix][b+iy],dp[0][a][b]+iz);  }  }  }  }  if(ix==0) temp=tempy;  }  memcpy(dp[0],dp[1],sizeof(dp[0]));  maxx+=tempx;  maxy+=temp;  }def-=c1*vx+c2*vy+c3*vz;  int ans=0;  for(int a=0;a<=maxx;a++)  {  for(int b=0;b<=maxy;b++)  {  if(dp[1][a][b]>=0)  {  rg(ans,a*vx+b*vy+dp[1][a][b]*vz+(def>0?def*min(a/c1,min(b/c2,dp[1][a][b]/c3)):0));  }  }  }  if(cs++) printf("\n");  printf("Case %d: %d\n",cs,ans);  }  }