Poj 2484 cow exhibition [deformation 0-1 backpack]

Question: poj 2484 cow Exhibition

Question: Give the nheaded ox. Each ox has a lucky Si and a smart ti. Now we need to select some cows to maximize the sum of the two values, provided that sum (SI) and sum (Ti) are both non-negative values.

Analysis: The data volume of this question is small, and it can be searched and trimmed.

Here we will talk about the idea of a 0-1 backpack. The obvious deformation of this question is that the item has two attribute values, and you must select the largest one.

So we can fix one value to find the maximum value of another. According to the idea of the 0-1 backpack, the state is defined: DP [I], which indicates loading some items to make sum (SI) the maximum sum (Ti) value.

Then the state transition equation is DP [I + Si] = max (DP [I + Si], DP [I] + Ti), which exactly represents the two choices of an item.

AC code:

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int N = 210000;const int pos = 100000;int dp[N];int main(){    int n;    while(~scanf("%d",&n))    {        memset(dp,-0x3f3f3f3f,sizeof(dp));        dp[pos] = 0;        int ma = pos,mi = pos;        for(int i=0;i<n;i++)        {            int s,t;            scanf("%d%d",&s,&t);            if(s>0){                for(int f=ma ;f>=mi;f--)                    dp[f+s] = max(dp[f+s],dp[f]+t);                ma+=s;            }            else{                for(int f = mi;f<=ma;f++)                    dp[f+s] = max(dp[f+s],dp[f]+t);                mi+=s;            }        }        int ans = 0;        for(int i=pos;i<=ma;i++)            if(dp[i]>=0)                ans = max(ans,i-pos+dp[i]);        printf("%d\n",ans);    }    return 0;}

Search timeout code:

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int N = 210000;const int pos = 100000;struct Node{    int s,t;};vector<Node> v;int cmp(Node a,Node b){    if(a.s!=b.s)        return a.s>b.s;    if(a.t!=b.t)        return a.t>b.t;}int ans;void dfs(int i,int sums,int sumt){    if(i==v.size())        return ;    if(sums>=0 && sumt>=0)        ans = max(ans,sums+sumt);    if(sums<0 && v[i].s<=0)        return ;    dfs(i+1,sums,sumt);    dfs(i+1,sums+v[i+1].s,sumt+v[i+1].t);}int main(){    int n;    int sums = 0,sumt = 0;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            int s,t;            scanf("%d%d",&s,&t);            if(s>=0 && t>=0)            {                sums+=s,sumt+=t;            }            else if(s<=0 && t<=0)                continue;            else                v.push_back((Node){s,t});        }        sort(v.begin(),v.end(),cmp);        ans = 0;        dfs(0,0,0);        dfs(0,v[0].s,v[0].t);        printf("%d\n",ans+sums+sumt);    }    return 0;}

Poj 2484 cow exhibition [deformation 0-1 backpack]