Zoj 2972 hurdles of 110 m

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 2972.

In the beginning, I thought it was a type of question similar to jiashuo. But the difference between this question and jiashuo is that each section has a speed value, which indicates that the state of jiashuo is a bit complicated.

Therefore, it is a little simpler.

DP [I] [J] indicates that the force of J is left when J is reached in segment I.

Therefore, it is clear that according to the question:

Stateful transition equation:

DP [I] [J] = min (DP [I-1] [J] + T2, DP [I] [J]);

If (j> = F1)
DP [I] [j-f1] = min (DP [I-1] [J] + T1, DP [I] [j-f1]);
Int temp = J + F2;
If (temp> m) temp = m;
DP [I] [temp] = min (DP [I] [temp], DP [I-1] [J] + T3 );

View code

# Include <iostream> # Include < String . H> # Include <Algorithm> # Include <Stdio. h> # Define Maxn 200 Using   Namespace  STD;  Const   Int INF = 10000  ;  Int  DP [maxn] [maxn];  Int  Main (){  Int  Test;  Int  N, m; Int  T1, T2, T3, F1, F2;  For (CIN> test; test -- ) {CIN >>N> M; fill (DP [  0 ], DP [n + 1 ], INF ); //  Attention should be paid to fill Functions DP [ 0 ] [M] = 0  ;  For ( Int I = 1 ; I <= N; I ++ ) {CIN > T1> T2> T3> F1> F2;  For ( Int J = m; j> = 0 ; J -- ) {DP [I] [J] = Min (DP [I- 1 ] [J] + T2, DP [I] [J]); //  Normal Mode                  If (J> = F1) DP [I] [J -F1] = min (DP [I-1 ] [J] + T1, DP [I] [j-f1]); //  Sufficient effort and maximum speed                  Int Temp = J + F2;  If (Temp> m) temp = m; //  The maximum value cannot be exceeded. DP [I] [temp] = min (DP [I] [temp], DP [I- 1 ] [J] + T3 ); //  Slowest effort  } Cout <* Min_element (DP [N], DP [N] + M + 1 ) <Endl;//  Pay attention to this function  }  Return   0  ;}