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dp...-------------------------------------------------------------------------------#include#include#include#include#include#define rep( i , n ) for( ... 阅读全文
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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
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直接floyd..----------------------------------------------------------------------------#include#include#include#include#define rep( i , n ) for( int i =... 阅读全文
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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
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有点类似背包 , 就是那样子搞...------------------------------------------------------------------------------------#include#include#include#include#define rep( i ,... 阅读全文
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直接floyd..-----------------------------------------------------------------------#include#include#include#include#include#define rep( i , n ) for( int ... 阅读全文
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dp , dp[ i ][ j ] = max( dp[ k ][ j - 1 ] ) + G[ i ][ j ] ( i - 1 0 )一开始没注意到要 dp[ k ][ j - 1 ] > 0 才能取 , 然后就WA 了2次...--------------------------------... 阅读全文
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从左往右,从右往左 扫两遍,单调栈维护...----------------------------------------------------------------------#include#include#include#include#include#define rep( i , n... 阅读全文
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MST...一开始没注意-1结果就WA了...----------------------------------------------------------------------------#include#include#include#include#include#include#de... 阅读全文
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背包dp..--------------------------------------------------------------------------------#include#include#include#include#define rep( i , n ) for( int i ... 阅读全文