HUST 1354 - Rubiks (DP)

1354 - Rubiks

时间限制：1秒 内存限制：64兆

452 次提交 102 次通过

题目描述

Isun is a genius. Not only he is an expert in algorithm, but also he plays damn-good in many funny games. Besides, he can recover a Rubik in 16 seconds or even less. The man is very crazy about Rubiks, and he has bought a lot of Rubiks. As we know, there are so many kinds of Rubiks in the world. Isun wants to buy the most valuable ones with his limited money. There are N kinds of Rubiks in all. Each of them has a price Pi(1<=i<=N) RMB and a value Vi(1<=i<=N). Isun will pay no more than M (RMB) in total. In addition, there are some Rubik families like “甲X” or “封X”. And a kind of Rubik belongs to one family at most. If Isun buys a group of them, the value of them as a family will increase. Can you get the largest value of the Rubiks that Isun can get with M (RMB). (Isun just buy one Rubik each kind at most)

输入

The input contains several test cases and is ended by EOF. Each test case begins with two integers: N (1<=N<=1000) and M (1<=M<=10000). The second line contains N integers representing the prices of the Rubiks. (1<=Pi<=10000) The third line contains N integers representing the value of the Rubiks. (1<=Vi<=10000) Then a line contains an integer G(0<=G<=15) representing the number of the Rubik families. Next G lines each with a start of an integer Si(1<=Si<=N) representing the number of Rubiks in the ith family. The following Si integers represent Rubik’s id (which start from 1 to N). And an integer Yi at the end means the value increased if you buy them all.(1<=Yi<=10000)

输出

There should be one line per test case containing the largest value.

样例输入

4 104 5 3 61 2 100 20012 1 2 330

样例输出

333

动态规划

#include <iostream>#include <string.h>#include <stdlib.h>#include <math.h>#include <algorithm>#include <stdio.h>using namespace std;int v[1020];int w[1020];int v2[20];int w2[20];int dp[10005];int bp[10005];int tag[1005];int b[20][1005];int n,m;int g;int s[20],sv;int a;int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%d",&w[i]);        for(int i=1;i<=n;i++)            scanf("%d",&v[i]);        scanf("%d",&g);        memset(tag,0,sizeof(tag));        for(int i=1;i<=g;i++)        {            scanf("%d",&s[i]);            int value=0;int weight=0;            for(int j=1;j<=s[i];j++)            {                scanf("%d",&b[i][j]);                tag[b[i][j]]=i;                value+=v[b[i][j]];                weight+=w[b[i][j]];            }            scanf("%d",&a);            value+=a;            w2[i]=weight;            v2[i]=value;        }        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            if(!tag[i])            {                for(int j=m;j>=w[i];j--)                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);            }        }        for(int i=1;i<=g;i++)        {            memcpy(bp, dp, sizeof(dp));            for(int j=m;j>=w2[i];j--)                    bp[j]=max(bp[j],bp[j-w2[i]]+v2[i]);                for(int k=1;k<=s[i];k++)                {                    for(int j=m;j>=w[b[i][k]];j--)                        dp[j]=max(dp[j],dp[j-w[b[i][k]]]+v[b[i][k]]);                }             for(int j=1;j<=m;j++)                dp[j]=max(dp[j],bp[j]);                    }        printf("%d\n",dp[m]);    }    return 0;}