[usaco3.3.2]shopping
题目传送门：http://www.nocow.cn/index.php/Translate:USACO/shopping
这道题我用dp做，把每一种优惠方案当作都有5件物品，没有的物品记为0，但写的很乱，所以直接看dp那里就好了
1 /* 2 ID:abc31261 3 LANG:C++ 4 TASK:shopping 5 */ 6 7 #include<cstdio> 8 #include<cstring> 9 #include<iostream> 10 using namespace std; 11 const int maxn=111,maxm=10; 12 int a[maxm],n[maxm],c[maxn][maxm],f[maxm][maxm][maxm][maxm][maxm],k[maxn][maxm],p1[maxn],yuan[maxn]; // yuan数组表示原价 13 int main() 14 { 15 int i,j,l,p,q,i1,i2,s,m; 16 freopen("shopping.in","r",stdin); 17 freopen("shopping.out","w",stdout); 18 scanf("%d",&s); 19 memset(f,0x7f,sizeof(f)); 20 memset(k,0,sizeof(k)); 21 memset(a,0,sizeof(a)); 22 f[0][0][0][0][0]=0; 23 for (i=1;i<=s;i++) 24 { 25 scanf("%d",&n[i]); 26 for (j=1;j<=n[i];j++)scanf("%d%d",&c[i][j],&k[i][j]); 27 scanf("%d",&p1[i]); 28 } 29 scanf("%d",&m); 30 for (i=1;i<=m;i++) 31 { 32 scanf("%d%d%d",&j,&a[i],&yuan[i]); 33 for (l=1;l<=s;l++) 34 for (p=1;p<=n[l];p++) 35 if (c[l][p]==j) 36 { 37 swap(c[l][p],c[l][i]); 38 swap(k[l][p],k[l][i]); 39 } 40 } 41 for (i=0;i<=a[1];i++) 42 for (j=0;j<=a[2];j++) 43 for (p=0;p<=a[3];p++) 44 for (q=0;q<=a[4];q++) 45 for (l=0;l<=a[5];l++) 46 { 47 if (i>0)f[i][j][p][q][l]=min(f[i][j][p][q][l],f[i-1][j][p][q][l]+yuan[1]); // dp方程 48 if (j>0)f[i][j][p][q][l]=min(f[i][j][p][q][l],f[i][j-1][p][q][l]+yuan[2]); 49 if (p>0)f[i][j][p][q][l]=min(f[i][j][p][q][l],f[i][j][p-1][q][l]+yuan[3]); 50 if (q>0)f[i][j][p][q][l]=min(f[i][j][p][q][l],f[i][j][p][q-1][l]+yuan[4]); 51 if (l>0)f[i][j][p][q][l]=min(f[i][j][p][q][l],f[i][j][p][q][l-1]+yuan[5]); 52 for (i1=1;i1<=s;i1++) //枚举所有优惠方案 53 if (i>=k[i1][1] && j>=k[i1][2] && p>=k[i1][3] && q>=k[i1][4] && l>=k[i1][5]) 54 f[i][j][p][q][l]=min(f[i][j][p][q][l], 55 f[i-k[i1][1]][j-k[i1][2]][p-k[i1][3]][q-k[i1][4]][l-k[i1][5]]+p1[i1]); 56 } 57 cout<<f[a[1]][a[2]][a[3]][a[4]][a[5]]<<endl; 58 return 0; 59 }
