HDU-1506 largest rectangle in a histogram Dynamic Programming

Attach the TleCodeFirst, discretization the height, and then solving the DP height. The complexity is too high.

The Code is as follows:

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# Include <cstdlib> # Include <Cstdio> # Include <Cstring> # Include <Map> # Include <Iostream> # Include <Algorithm> # Define Maxn 100005 Using   Namespace  STD; Int  N, CNT;  Int  SEQ [maxn], CSeq [maxn];  Long   Long  DP [maxn]; Map < Int , Int > MP, RMP;  Void Getint ( Int & C ){  Char  CHR; While (CHR = getchar (), CHR < '  0  ' | CHR> '  9  '  ); C = CHR- '  0  '  ;  While (CHR = getchar (), CHR> = '  0  ' & CHR <= '  9  '  ) {C = C * 10 + Chr- '  0  '  ;} Inline  Long   Long  DP (){  Long   Long Max = 0  ; For ( Int I = 0 ; I <n; ++ I ){  For ( Int J = 1 ; J <= MP [seq [I]; ++ J) {DP [J] + = RMP [J]; max = Max (max, DP [J]);}  For ( Int J = MP [seq [I] + 1 ; J <= CNT; ++ J) {DP [J] = 0  ;}}  Return  Max ;}  Int  Main (){  Int  Max;  While (Scanf ( "  % D  " ,& N), n) {CNT = 0 ; Memset (DP,  0 , Sizeof  (DP); MP. Clear (), RMP. Clear ();  For ( Int I = 0 ; I <n; ++ I) {getint (SEQ [I]); CSeq [I] = SEQ [I];} Sort (CSeq, CSeq + N );  For ( Int I =0 ; I <n; ++ I ){  If (MP. Count (CSeq [I]) = 0  ) {MP [CSeq [I] = ++ CNT; RMP [CNT] = CSeq [I] ;}} printf (  "  % I64d \ n  "  , Dp ());}  Return   0  ;} 

 

Later, the state is constructed based on a more sharp solution. L [I], R [I], and the two arrays are maintained for the purpose. It indicates the leftmost and rightmost aspects of the current height.

The Code is as follows:

# Include <cstdlib> # Include <Cstdio> # Include <Cstring> # Include <Iostream> # Define Maxn 100005 Using   Namespace  STD;  Int  N, seq [maxn], L [maxn], R [maxn]; inline  Long   Long Max (Long   Long X, Long   Long  Y ){  Return X> Y? X: Y ;}  Long   Long  DP (){  Long   Long Max = 0  ;  For ( Int I =2 ; I <= N; ++ I ){  While (SEQ [L [I]- 1 ]> = SEQ [I]) {L [I] = L [L [I]- 1  ];  If (L [I] <= 0 ) Break ; //  This sentence is not added before, TLE  }}  For (Int I = N- 1 ; I> = 1 ;-- I ){  While (SEQ [R [I] + 1 ]> = SEQ [I]) {R [I] = R [R [I] + 1  ];  If (R [I]> = N) Break  ;}}  For ( Int I = 1 ; I <= N; ++ I) {max = Max (max, (long) (1) * seq [I] * (R [I]-l [I] + 1  ));}  Return  Max ;}  Int  Main (){  While (Scanf ( "  % D  " ,& N), n ){  For (Int I = 1 ; I <= N; ++ I) {scanf (  "  % D  " ,& SEQ [I]); L [I] = R [I] = I;} printf (  "  % I64d \ n  "  , Dp ());}  Return   0  ;}