poj2392 1742多重背包

多重背包就是每组物品只能取有限多个

所以dp是要根据情况加一个if判断

首先按最大高度进行升序排序

另外加一个num 记录该数值已经取了多少个当前类型的block

#include<iostream>
#include <string>
#include<vector>
#include<algorithm>
#include<set>
#include<fstream>
#include<cmath>
#include<queue>
using namespace std;
#define lch(i) ((i)<<1)
#define rch(i) ((i)<<1|1)
#define sqr(i) ((i)*(i))
#define pii pair<int,int>
#define mp make_pair
#define FOR(i,b,e) for(int i=b;i<=e;i++)
#define FORE(i,b,e) for(int i=b;i>=e;i--)
#define ms(a)   memset(a,0,sizeof(a))
const int maxnum =40005;
const int INF = 25000005;
int tol,n,m;
int dp[maxnum];

struct node
{
int h,top,num;
node(){}
node(int h1,int top1,int num1):h(h1),top(top1),num(num1){}
};

int cmp(node a,node b){
return a.top<b.top;
}

int main()
{
//fstream fin("G:/1.txt");
cin>>n;
//fin>>n;
vector<node> vec(n,node(0,0,0));
int u,v,w;
FOR(i,1,n){
scanf("%d%d%d",&u,&v,&w);
//fin>>u>>v>>w;
vec[i-1].h=u;
vec[i-1].top=v;
vec[i-1].num=w;
}
sort(vec.begin(),vec.end(),cmp);
dp[0]=1;
int num[40005];
FOR(i,1,n){
ms(num);
FOR(j,vec[i-1].h,vec[i-1].top){
if(!dp[j]&&dp[j-vec[i-1].h]&&num[j-vec[i-1].h]<vec[i-1].num){
dp[j]=1;
num[j]=num[j-vec[i-1].h]+1;
}
}
}
int sum=vec[n-1].top;
while(dp[sum]==0) sum--;
cout<<sum;
return 0;
}

1742 差不多

#include<iostream>
#include <string>
#include<vector>
#include<algorithm>
#include<set>
#include<fstream>
#include<cmath>
#include<queue>
using namespace std;
#define lch(i) ((i)<<1)
#define rch(i) ((i)<<1|1)
#define sqr(i) ((i)*(i))
#define pii pair<int,int>
#define mp make_pair
#define FOR(i,b,e) for(int i=b;i<=e;i++)
#define FORE(i,b,e) for(int i=b;i>=e;i--)
#define ms(a) memset(a,0,sizeof(a))
const int maxnum =100005;
const int INF = 25000005;
int tol,n,m;
int dp[maxnum];

struct node
{
int v,num;
node(){}
node(int v1,int num1):v(v1),num(num1){}
}tp[maxnum];

int main()
{
//fstream fin("G:/1.txt");
//while(fin>>n>>m){
while(1){
scanf("%d%d",&n,&m);
if(n==0&&m==0)break;
ms(dp);ms(tp);
FOR(i,1,n){
scanf("%d",&tp[i].v);
//fin>>tp[i].v;
}
FOR(i,1,n){
scanf("%d",&tp[i].num);
//fin>>tp[i].num;
}

int num[maxnum];
dp[0]=1;
FOR(i,1,n){
ms(num);
FOR(j,tp[i].v,m){
if(!dp[j]&&dp[j-tp[i].v]&&num[j-tp[i].v]<tp[i].num){
dp[j]=1;
num[j]=num[j-tp[i].v]+1;
}
}
}
int sum=0;
FOR(i,1,m){
if(dp[i])sum++;
}
printf("%d\n",sum);
}
return 0;
}