zoj1013 great equipment

範例輸入

3

1 1 3

5 6 10

2 1 2

1 1 1 50

1 1

5 6

2 1

範例輸出

50

典型的DP最優填表是二維數組，行表示階段，列表示狀態，繼而表格內資料即為該階段該狀態下的最優值。該題每一階段需要用二維數組表示，每遍曆完一個狀態就對數組更新一次，這樣數組內始終是目前狀態下的最優值。

#include<iostream>using namespace std;#define NUM 501int trade[NUM][NUM],carry[NUM][NUM];int min(int x,int y){if(x>y) return y;else return x;}void main(){int w[4],s[4],d[4],i,n;cin>>n;for(i=1;i<=3;i++)cin>>w[i]>>s[i]>>d[i];int c1,c2,c3,d4;cin>>c1>>c2>>c3>>d4;d4-=c1*d[1]+c2*d[2]+c3*d[3];memset(carry,-1,sizeof(carry));carry[0][0]=0;  //初始化什麼都不裝int row=0,col=0,j,k,ja,ka;for(i=0;i<n;i++){int weight,size;cin>>weight>>size;memset(trade,-1,sizeof(trade));int newrow=row,newcol=col;   //追蹤有效地行數和列數，只對有效地行列操作int weight1,size1,weight2,size2;for(j=0;j<=row;j++)for(k=0;k<=col;k++)if(carry[j][k]>=0)  //原來存在該種裝備組合情況，在該組合基礎上對當前馬車運載組合進行枚舉for(ja=j,weight1=size1=0;(weight1<=weight&&size1<=size);weight1+=w[1],size1+=s[1],ja++)for(ka=k,weight2=weight1,size2=size1;(weight2<=weight&&size2<=size);weight2+=w[2],size2+=s[2],ka++){if(newrow<ja) newrow=ja;if(newcol<ka) newcol=ka;int bootweight=(weight-weight2)/w[3];int bootsize=(size-size2)/s[3];if(bootweight>bootsize)bootweight=bootsize;bootweight+=carry[j][k];if(trade[ja][ka]<bootweight)trade[ja][ka]=bootweight;}memcpy(carry,trade,sizeof(trade));row=newrow;col=newcol;}int ibest=0;     //搜尋carry[j][k]獲得最優值for(j=0;j<=row;j++)for(k=0;k<=col;k++)if(carry[j][k]>=0){int defend=j*d[1]+k*d[2]+carry[j][k]*d[3];int helms=j/c1;int armors=k/c2;int boots=carry[j][k]/c3;if(d4>0)defend+=d4*min(helms,min(armors,boots));if(ibest<defend)ibest=defend;}cout<<ibest<<endl;}