ZOJ 3211 Dream City【dp】

Dream City
Time Limit: 1 Second      Memory Limit: 32768 KB

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 ai coins on it (i=1, 2, 3...n). Surprisingly, each tree i can grow bi 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, ai and bi (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 ai. (0 < ai <= 100, i=1, 2, 3...n) The third line of each test case also contains n positive
integers separated by a space, indicating bi. (0 < bi <= 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.

 思路：dp[i][j]表示i件物品选择j件的最大价值。

状态转移方程:dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]+val+(j-1)*xk) (val为初始值，xk为每天增加的价值)

注意：增长率小的物品，如果在某个阶段不选，那么在以后都不会再选。

#include<cstdio>
#include<math.h>
#include<cstring>
#include<climits>
#include<string>
#include<queue>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<climits>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
using namespace std;
#define rep(i,j,k)for(i=j;i<k;i++)
#define per(i,j,k)for(i=j;i>k;i--)
#define MS(x,y)memset(x,y,sizeof(x))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ll long long
#define abs(x) (x>0?x:-x)
const int INF=0x7ffffff;
const ll MAX=1e18;

int i,j,k,n,m;
int dp[300][300];
struct Node
{
int a,b;
}tree[300];

bool cmp(Node x,Node y)
{
return x.b<y.b;
}

int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)scanf("%d",&tree[i].a);
for(i=1;i<=n;i++)scanf("%d",&tree[i].b);
sort(tree+1,tree+n+1,cmp);
for(i=1;i<=n;i++)
for(j=1;j<=m;j++){
dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+tree[i].a+tree[i].b*(j-1));
}
printf("%d\n",dp
[m]);
}
return 0;
}


Dream City
Time Limit: 1 Second      Memory Limit: 32768 KB
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 ai coins on it (i=1, 2, 3...n). Surprisingly, each tree i can grow bi 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, ai and bi (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 ai. (0 < ai <= 100, i=1, 2, 3...n) The third line of each test case also contains n positive
integers separated by a space, indicating bi. (0 < bi <= 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.