[Uva12260]Free Goodies(dp+贪心)

解题关键：先对p进行排序，消除p的影响，然后对w进行01背包即可。注意p对w的约束。j<=(cur+1)/2

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<iostream>
using namespace std;
typedef long long ll;
struct node{
    int p,w;
    bool operator<(const node& a)const{
        return p>a.p||(p==a.p&&w<a.w);
    }
}nod[];
int dp[][],cost[][],t,n;
string s;
int main(){
    cin>>t;
    while(t--){
        int sum=;
        memset(dp,,sizeof dp);
        memset(cost,,sizeof cost);
        cin>>n>>s;
        for(int i=;i<=n;i++) cin>>nod[i].p>>nod[i].w,sum+=nod[i].p;
        sort(nod+,nod+n+);
        int cur=;
        for(int i=s[]=='P'?:;i<=n;i++){
            cur++;
            for(int j=;j<=(cur+)/;j++){
                dp[i][j]=dp[i-][j];
                cost[i][j]=cost[i-][j];
                if(j!=&&!dp[i-][j-]) continue;
                if(dp[i][j]<dp[i-][j-]+nod[i].w){
                    dp[i][j]=dp[i-][j-]+nod[i].w;
                    cost[i][j]=cost[i-][j-]+nod[i].p;
                }else if(dp[i][j]==dp[i-][j-]+nod[i].w){
                    cost[i][j]=min(cost[i][j],cost[i-][j-]+nod[i].p);
                }
            }
        }
        cout<<sum-cost[n][(cur+)/]<<" "<<dp[n][(cur+)/]<<"\n";
    }
    return ;
}

优化之后：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<iostream>
#define inf 1<<30
using namespace std;
typedef long long ll;
struct node{
    int p,w;
    bool operator<(const node& a)const{
        return p>a.p||(p==a.p&&w<a.w);
    }
}nod[];
int dp[],cost[],t,n;
string s;
int main(){
    cin>>t;
    while(t--){
        int sum=;
        memset(dp,,sizeof dp);
        fill(cost,cost+,inf);
        cin>>n>>s;
        for(int i=;i<=n;i++) cin>>nod[i].p>>nod[i].w,sum+=nod[i].p;
        sort(nod+,nod+n+);
        int cur=;
        cost[]=;
        for(int i=s[]=='P'?:;i<=n;i++){
            cur++;
            for(int j=(cur+)/;j>=;j--){//背包容量，每个物品的容量是1
                if(dp[j-]+nod[i].w>dp[j]){
                    dp[j]=dp[j-]+nod[i].w;
                    cost[j]=cost[j-]+nod[i].p;
                }else if(dp[j-]+nod[i].w==dp[j]){
                    cost[j]=min(cost[j],cost[j-]+nod[i].p);
                }
            }
        }

        cout<<sum-cost[(cur+)/]<<" "<<dp[(cur+)/]<<"\n";
    }
    return ;
}