[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[1002];
int dp[1002][1002],cost[1002][1002],t,n;
string s;
int main(){
    cin>>t;
    while(t--){
        int sum=0;
        memset(dp,0,sizeof dp);
        memset(cost,0,sizeof cost);
        cin>>n>>s;
        for(int i=1;i<=n;i++) cin>>nod[i].p>>nod[i].w,sum+=nod[i].p;
        sort(nod+1,nod+n+1);
        int cur=0;
        for(int i=s[0]=='P'?2:1;i<=n;i++){
            cur++;
            for(int j=1;j<=(cur+1)/2;j++){
                dp[i][j]=dp[i-1][j];
                cost[i][j]=cost[i-1][j];
                if(j!=1&&!dp[i-1][j-1]) continue;
                if(dp[i][j]<dp[i-1][j-1]+nod[i].w){
                    dp[i][j]=dp[i-1][j-1]+nod[i].w;
                    cost[i][j]=cost[i-1][j-1]+nod[i].p;
                }else if(dp[i][j]==dp[i-1][j-1]+nod[i].w){
                    cost[i][j]=min(cost[i][j],cost[i-1][j-1]+nod[i].p);
                }
            }
        }
        cout<<sum-cost[n][(cur+1)/2]<<" "<<dp[n][(cur+1)/2]<<"
";
    }
    return 0;
}

 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[1002];
int dp[1002],cost[1002],t,n;
string s;
int main(){
    cin>>t;
    while(t--){
        int sum=0;
        memset(dp,0,sizeof dp);
        fill(cost,cost+1001,inf);
        cin>>n>>s;
        for(int i=1;i<=n;i++) cin>>nod[i].p>>nod[i].w,sum+=nod[i].p;
        sort(nod+1,nod+n+1);
        int cur=0;
        cost[0]=0;
        for(int i=s[0]=='P'?2:1;i<=n;i++){
            cur++;
            for(int j=(cur+1)/2;j>=1;j--){//背包容量，每个物品的容量是1
                if(dp[j-1]+nod[i].w>dp[j]){
                    dp[j]=dp[j-1]+nod[i].w;
                    cost[j]=cost[j-1]+nod[i].p;
                }else if(dp[j-1]+nod[i].w==dp[j]){
                    cost[j]=min(cost[j],cost[j-1]+nod[i].p);
                }
            }
        }
        
        cout<<sum-cost[(cur+1)/2]<<" "<<dp[(cur+1)/2]<<"
";
    }
    return 0;
}