第六周總結修正版

一、知識總結

(1)滾動陣列

二維dp只需兩行或兩列時，用[1][0]來輪換表示[j],

[j-1]，從而降低空間複雜度。

j1=1;dp[i][j11]借助是j1的值在1,0間反覆換。

(2)省時輸入

scanf輸入比cin輸入節省時間，但輸入多種型別

時cin更方便。所以遇到輸入量特別大時，如果

想用cin，需要先做處理。

//**

int main

//二、重要**樣例

g-ga project manager wants to determine the

number of the workers needed in every

month. he does know the minimal number

of the workers needed in each month. when

he hires or fires a worker, there will be some

extra cost. once a worker is hired, he will get

the salary even if he is not working. the

manager knows the costs of hiring a worker,

firing a worker, and the salary of a worker.

then the manager will confront such a

problem: how many workers he will hire or

fire each month in order to keep the lowest

total cost of the project.

input

the input may contain several data sets.

each data set contains three lines. first line

contains the months of the project planed to

use which is no more than 12. the second

line contains the cost of hiring a worker, the

amount of the salary, the cost of firing a

worker. the third line contains several

numbers, which represent the minimal

number of the workers needed each month.

the input is terminated by line containing a

single 『0』.

output

the output contains one line. the minimal

total cost of the project.

sample input

34 5 6

10 9 11

0sample output

199**:

#include
using namespace std;
const
int inf=
0x3f3f3f3f
;int dp[20]
[100000];
intmain()
,maxn=
0,sum=inf;
scanf
("%d%d%d"
,&hire,
&salary,
&fire)
;for
(int i=
1;i<=n;i++
)for
(int i=
1;i<=n;i++)}
for(
int i=month[1]
;i<=maxn;i++
) 
dp[1]
[i]=
(hire+salary)
*i;for
(int i=
2;i<=n;i++)}
}for
(int i=month[n]
;i<=maxn;i++
) sum=
min(sum,dp[n]
[i])
;printf
("%d\n"
,sum)
;return0;
}

j-j

now here is a matrix with letter 『a』,『b』,『c』,『w』,『x』,『y』,『z』

and you can change 『w』 to 『a』 or 『b』, change 『x』 to

『b』 or 『c』, change 『y』 to 『a』 or 『c』, and change 『z』 to

『a』, 『b』 or 『c』. after you changed it, what』s the

largest submatrix with the same letters you can

make?

input

the input contains multiple test cases. each

test case begins with m and n (1 ≤ m, n ≤

1000) on line.

then come the elements of a matrix in

row-major order on m lines each with n

letters. the input ends once eof is met.

output

for each test case, output one line containing

the number of elements of the largest

submatrix of all same letters.

sample input

2 4abcw

wxyz

sample output

#include
#include
#include
#include
using namespace std;
const
int maxn=
1e3+10;
char g[maxn]
[maxn]
;int dp[maxn]
[maxn]
,area,n,m;
void
get_area()
for(
int j=m-
1; j>=
1; j--
)for
(int j=
1; j<=m; j++)}
}int
main()
area=
0; memset (dp,
0,sizeof (dp));
for(
int j=
1; j<=m; j++)}
get_area()
; 
memset (dp,
0,sizeof (dp));
for(
int j=
1; j<=m; j++)}
get_area()
; 
memset (dp,
0,sizeof (dp));
for(
int j=
1; j<=m; j++)}
get_area()
;printf
("%d\n"
,area);}
return0;
}

三、做題感想

vjudge有幾個題，我抄的別人的狀態轉移方程。在草稿紙上演算發現dp陣列不一定要必須連續賦值。像g-g題的狀態轉移方程有些行只有一列被賦值，其餘列都是初值0，而且存在對同乙個變數多次賦值的情況。另一情況是動態規劃中也有題目要用輔助函式，麻煩的是像j-j一樣要多次用到輔助函式。這幾天做動態規劃，覺得自己在思維方面很差，許多狀態轉移方程都是搜出來的，最核心的部分自己不會，一定有後患。我應該擠時間把線性 dp的例題做一遍。ppt有較為詳細的思路介紹，有些題也是經典題目，先掌握最大欄位和，分配等基礎問題才會有個好底子。

第六周總結 修正版

第六周總結

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第六周總結修正版