ZOJ3211-Dream City(贪心思想+变形的01背包）

Dream City

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3211

Description

JAVAMAN is visiting Dream City and he sees a yard of gold coin trees. There are n trees in the yard. Let's call them tree 1, tree 2 ...and tree n. At the first day, each tree i has a_i coins on it (i=1, 2, 3...n). Surprisingly, each tree i can grow b_i new coins each day if it is not cut down. From the first day, JAVAMAN can choose to cut down one tree each day to get all the coins on it. Since he can stay in the Dream City for at most m days, he can cut down at most m trees in all and if he decides not to cut one day, he cannot cut any trees later. (In other words, he can only cut down trees for consecutive m or less days from the first day!)

Given n, m, a_i and b_i (i=1, 2, 3...n), calculate the maximum number of gold coins JAVAMAN can get.

 

 

Input

 

There are multiple test cases. The first line of input contains an integer T (T <= 200) indicates the number of test cases. Then T test cases follow.

Each test case contains 3 lines: The first line of each test case contains 2 positive integers n and m (0 < m <= n <= 250) separated by a space. The second line of each test case contains n positive integers separated by a space, indicating a_i. (0 < a_i <= 100, i=1, 2, 3...n) The third line of each test case also contains n positive integers separated by a space, indicating b_i. (0 < b_i <= 100, i=1, 2, 3...n)

 

Output

 

For each test case, output the result in a single line.

 

Sample Input

 

 

2
2 1
10 10
1 1
2 2
8 10
2 3

 

 

Sample Output

 

 

10
21

 

Hints:
Test case 1: JAVAMAN just cut tree 1 to get 10 gold coins at the first day.
Test case 2: JAVAMAN cut tree 1 at the first day and tree 2 at the second day to get 8 + 10 + 3 = 21 gold coins in all.

 

题目意思：
有n颗树，每棵树都有固定的初始值,和有增长值，
给出m天，每天只能砍一棵树，那么对与第j天，
如果选择砍掉第i棵树，
那么收获的硬币就是初始值+增长值*(j-1)，
怎样砍才能使收获的硬币最多

解题思路：增长快的要放在后面砍会比较有利。
在做背包前，要先排个序，
可将问题转化为01背包问题。
dp[i][j]表示前i课树，第j天所能取到的最大值。
状态转移方程 :
dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+p[i].v1+p[i].v2*(j-1));

 

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define eps 10e-6
#define INF 999999999
using namespace std;
#define max_v 255
struct node
{
    int v1;
    int v2;
}p[max_v];
bool cmp(node a,node b)
{
    return a.v2<b.v2;
}
int dp[max_v][max_v];
int main()
{
    int t;
    int n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);

        for(int i=1;i<=n;i++)
        {
            scanf("%d",&p[i].v1);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&p[i].v2);
        }

        sort(p+1,p+1+n,cmp);

        memset(dp,0,sizeof(dp));

        int ans=-INF;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+p[i].v1+p[i].v2*(j-1));
                ans=max(ans,dp[i][j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}
/*

题目意思：
有n颗树，每棵树都有固定的初始值,和有增长值，
给出m天，每天只能砍一棵树，那么对与第j天，
如果选择砍掉第i棵树，
那么收获的硬币就是初始值+增长值*(j-1)，
怎样砍才能使收获的硬币最多

解题思路：增长快的要放在后面砍会比较有利。
在做背包前，要先排个序，
可将问题转化为01背包问题。
dp[i][j]表示前i课树，第j天所能取到的最大值。
状态转移方程 :
dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+p[i].v1+p[i].v2*(j-1));

*/