zoj2156Charlie's Change 多重背包可行性问题+记录路径

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zoj2156

用一个结构体pre数组来记录前一步的信息

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int C,n[30],v[5]={0,1,5,10,25};int dp[15000];struct PRE{int ID,NUM,P;} pre[15000];int main(){int i,j,k,x;while(1){scanf("%d%d%d%d%d",&C,&n[1],&n[2],&n[3],&n[4]);if(!(C|n[1]|n[2]|n[3]|n[4]))break;for(i=0;i<=C;i++){dp[i]=-1;pre[i].ID=i;pre[i].NUM=pre[i].P=0;}dp[0]=0;for(i=1;i<=4;i++){x=1;while(n[i]>0){if(x>n[i])x=n[i];n[i]-=x;for(j=C;j>=x*v[i];j--)if(dp[j-x*v[i]]>=0&&dp[j]<dp[j-x*v[i]]+x){dp[j]=dp[j-x*v[i]]+x;pre[j].ID=j-x*v[i];pre[j].NUM=x;pre[j].P=v[i];}x*=2;}}//for(i=1;i<=C;i++)printf("%d ",pre[i].ID);printf("\n");//while(1);//for(i=1;i<=C;i++)printf("%d ",dp[i]);printf("\n");if(dp[C]>0){for(memset(n,0,sizeof(n)),x=C;x!=0;x=pre[x].ID)n[pre[x].P]+=pre[x].NUM;printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",n[1],n[5],n[10],n[25]);}elseprintf("Charlie cannot buy coffee.\n");}return 0;}