CF 3B. Lorry

CF 3B. LorryClick    to open the link

The meaning of the question: There are n types of items with a volume of 1 or 2. Your capacity is m. How many items of value can you get at most?

Idea: Greedy thinking coupled with the method of taking the ruler to obtain the greatest value, this is what I wrote after reading the blog of the big guy, so it is classified as a reprint.

Define a structure, record the volume w, the value v, and the subscript index. Then the weight of 1 is sorted in ascending order, and the weight of 2 is sorted in descending order, and then the ruler is taken.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std;
#define MAXN 100000+10
struct Node{
    int w,v,index;
 } node [MAXN];
 int n,m;
 int cmp(const Node &p,const Node &q){
     if(p.w!=q.w){
         return p.w<q.w;
     }else if(p.w==1){
         return p.v<q.v;
     }else
         return p.v>q.v;
 }

 int main(){
     while(~scanf("%d%d",&n,&m)){
         for(int i=1;i<=n;i++){
             int x,y;
             scanf("%d%d",&x,&y);
             node[i].index=i;
             node[i].v=y;
             x==1?node[i].w=1:node[i].w=2;
         }
         sort(node+1,node+n+1,cmp);
         int MAX=0,w1=0,v1=0,MAXi=1;
         for(int i=1,j=1;i<=n;i++){
             w1+=node[i].w;
             v1+=node[i].v;
             while(w1>m){
                 w1-=node[j].w;
                 v1-=node[j].v;
                 j++;//If the capacity is exceeded, subtract what was added last time,//
             }
             if(v1>MAX){MAX=v1;MAXi=j;}//Use Max to record the maximum total value//
         }
         printf("%d\n",MAX);
         w1=v1=0;
         for(int j=MAXi;j<=n;j++){
             w1+=node[j].w;
             v1+=node[j].v;
             if(w1>m)break;
             j==MAXi?printf("%d",node[j].index):printf(" %d",node[j].index);
         }
         puts("");
     }
     return 0;
 }