poj2392 1742多重背包

来源：互联网发布：mac炒股软件哪个好编辑：程序博客网时间：2024/05/17 07:44

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;}

0 0