Description
Each vase has a distinct characteristic (just like flowers do). Hence, putting a bunch of flowers in a vase results in a certain aesthetic value, expressed by an integer. The aesthetic values are presented in a table as shown below. Leaving a vase empty has an aesthetic value of 0.
V A S E S |
||||||
1 |
2 |
3 |
4 |
5 |
||
Bunches |
1 (azaleas) |
7 | 23 | -5 | -24 | 16 |
2 (begonias) |
5 | 21 | -4 | 10 | 23 | |
3 (carnations) |
-21 |
5 | -4 | -20 | 20 |
According to the table, azaleas, for example, would look great in vase 2, but they would look awful in vase 4.
To achieve the most pleasant effect you have to maximize the sum of aesthetic values for the arrangement while keeping the required ordering of the flowers. If more than one arrangement has the maximal sum value, any one of them will be acceptable. You have to produce exactly one arrangement.
Input
- The first line contains two numbers: F, V.
- The following F lines: Each of these lines contains V integers, so that Aij is given as the jth number on the (i+1)st line of the input file.
- 1 <= F <= 100 where F is the number of the bunches of flowers. The bunches are numbered 1 through F.
- F <= V <= 100 where V is the number of vases.
- -50 <= Aij <= 50 where Aij is the aesthetic value obtained by putting the flower bunch i into the vase j.
Output
Sample Input
3 5 7 23 -5 -24 16 5 21 -4 10 23 -21 5 -4 -20 20
Sample Output
53
题意:
f朵花放到v个瓶子里,每朵花放到每个瓶子都有相应的观赏值,一个瓶子只能放一朵花,求最大的观赏值。
用dp[i][j]表示i朵花放到j个瓶子里所得到的最大观赏值,可得状态转移方程:
dp[i][j] = dp[i][j-1]>dp[i-1][j-1]+a[i][j]?dp[i][j-1]:dp[i-1][j-1]+a[i][j] ;
代码:
#include<iostream>
#include<cstring>
using namespace std ;
int main(){
int f, v, i, j ;
int a[110][110] ;
int dp[110][110] ;
while(cin >> f >> v){
for(i=1; i<=f; i++)
for(j=1; j<=v; j++)
cin >> a[i][j] ;
memset(dp, 0, sizeof(dp)) ;
dp[1][1] = a[1][1] ;
for(i=2; i<=f; i++)
dp[1][i] = dp[1][i-1]>a[1][i]?dp[1][i-1]:a[1][i] ;
for(i=1; i<=f; i++){
dp[i][i] = dp[i-1][i-1] + a[i][i] ;
for(j=i+1; j<=v; j++)
dp[i][j] = dp[i][j-1]>dp[i-1][j-1]+a[i][j]?dp[i][j-1]:dp[i-1][j-1]+a[i][j] ;
}
cout << dp[f][v] << endl ;
}
return 0 ;
}