Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 660 Accepted Submission(s): 196
#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
struct Cake{
int energy,siz,amont;
}cakes[220];
struct Truck{
int siz,cost,num;
}trucks[220];
int dp1[55000],dp2[55000];
void ZeroOnePack(int cost,int weight,int V,int *dp,int typ){
if(typ==1)
for(int i=V;i>=cost;i--){
dp[i]=min(dp[i],dp[i-cost]+weight);
}
else
for(int i=V;i>=cost;i--){
dp[i]=max(dp[i],dp[i-cost]+weight);
}
}
void CompletePack(int cost,int weight,int V,int *dp,int typ){
if(typ==1)
for(int i=cost;i<=V;i++){
dp[i]=min(dp[i],dp[i-cost]+weight);
}
else
for(int i=cost;i<=V;i++){
dp[i]=max(dp[i],dp[i-cost]+weight);
}
}
void MultiplePack(int cost,int weight,int amount,int V,int *d,int typ){
if(cost*amount>=V){
CompletePack(cost,weight,V,d,typ);
return ;
}
int k=1;
while(amount>k){
ZeroOnePack(cost*k,weight*k,V,d,typ);
amount-=k;
k*=2;
}
ZeroOnePack(cost*amount,weight*amount,V,d,typ);
}
int main(){
int T,n,m,p;
scanf("%d",&T);
while(T--){
memset(dp1,INF,sizeof(dp1));
memset(dp2,0,sizeof(dp2));
dp1[0]=0;
int vv=0,cc=0;
scanf("%d%d%d",&n,&m,&p);
for(int i=1;i<=n;i++){
scanf("%d%d%d",&cakes[i].energy,&cakes[i].siz,&cakes[i].amont);
vv+=cakes[i].energy*cakes[i].amont;
}
for(int i=1;i<=m;i++){
scanf("%d%d%d",&trucks[i].siz,&trucks[i].cost,&trucks[i].num);
cc+=trucks[i].cost*trucks[i].num;
}
vv=min(50000,vv);
for(int i=1;i<=n;i++){
MultiplePack(cakes[i].energy,cakes[i].siz,cakes[i].amont,vv,dp1,1);
}
cc=min(cc,50000);
for(int i=1;i<=m;i++){
MultiplePack(trucks[i].cost,trucks[i].siz,trucks[i].num,cc,dp2,2);
}
int pos=0;
int tmp=INF;
for(int i=p;i<=vv;i++){
tmp=min(dp1[i],tmp);
}
for(int i=1;i<=cc;i++){
if(dp2[i]>=tmp){
pos=i; break;
}
}
if(pos==0){
printf("TAT\n");
}else{
printf("%d\n",pos);
}
}
return 0;
}
/*
555
5 3 34
1 4 1
9 4 2
5 3 3
1 3 3
5 3 2
3 4 5
6 7 5
5 3 8
*/
HDU 5445——Food Problem——————【多重背包】
原文:http://www.cnblogs.com/chengsheng/p/4834928.html