摘要:
dp[ i ][ j ] = max( dp[ i - 1 ][ k ] + w[ i ][ j - k ] ) ( 0 #include#include#include#include#define rep( i , n ) for( int i = 0 ; i > n >> m >> T; ... 阅读全文
摘要:
直接floyd..----------------------------------------------------------------------------#include#include#include#include#define rep( i , n ) for( int i =... 阅读全文
摘要:
dpdp[ i ] 表示第 i 个不选 , 前 i 个的选择合法的最小损失 , dp[ i ] = min( dp[ j ] ) ( max( 0 , i - 1 - k ) #include#include#include#include#define rep( i , n ) for( int ... 阅读全文
摘要:
有点类似背包 , 就是那样子搞...------------------------------------------------------------------------------------#include#include#include#include#define rep( i ,... 阅读全文