Poj 1042 (Greedy)

Http://poj.org/problem? Id = 1042

The first question made during the winter vacation was definitely abused .. Lili's wa had no more than 20 times, and finally there was no way to let others find out the error! Sad reminder...

 

This question is greedy, enumerative to the end of the island I, and the time spent on the island is reduced from the total time. Then greedy for the maximum value in I Island.

There are two errors:

1. The question requires extra time to be placed on the first Island. I added cur = 0 to the selected max, and added the time directly to the first Island.

In fact, this does not need to be processed separately. cur = 0 has been made before greed. If max <0. Fish, cur will not change. One click

2. The order of resetting and decreasing when the number of fish is negative. After clearing, I make cur = 0 and then decrease f [0.

In this case, the operation can be negative for the fish, but it cannot be done if the fish is negative after the decrease, which may cause the island selection error in the greedy process.

 

# Include <iostream>
# Include <fstream>
# Include <cstring>
# Include <cstdlib>
# Define max (A, B) A> B? A: B
Using namespace STD;

Struct f {
Int fish;
Int DESC;
Int walk;
};
F [30];
F ftemp [30];
Int time [30] [30];
Int sum [30];

Int main (){
Int N, H, total, I, j, Max, maxf, cur, temp;
// Fstream CIN ("in. dat", IOS: In );
While (~ Scanf ("% d", & N ){
Scanf ("% d", & H );
For (I = 0; I <n; I ++)
Scanf ("% d", & F [I]. Fish );
For (I = 0; I <n; I ++)
Scanf ("% d", & F [I]. DESC );
For (I = 1; I <n; I ++)
Scanf ("% d", & F [I]. Walk );
Total = 12 * h;
F [0]. Walk = 0;
Memset (time, 0, sizeof (time ));
Memset (sum, 0, sizeof (SUM ));
For (I = 0; I <n; I ++) {// enumeration ends at Lake I
Total-= f [I]. Walk;
Temp = total;
Memcpy (ftemp, F, (I + 1) * sizeof (f ));
While (temp> 0) {// greedy
Maxf = ftemp [0]. fish;
Cur = 0; // each time the greedy start mark the position cur = 0 ensures that the remaining time belongs
For (j = 1; j <= I; j ++ ){
If (maxf <ftemp [J]. Fish ){
Maxf = ftemp [J]. fish;
Cur = J;
}
}
Sum [I] + = ftemp [cur]. fish;
Ftemp [cur]. Fish-= ftemp [cur]. DESC;
// Here we need to reduce the number before solving the non-positive situation.
If (ftemp [cur]. Fish <0)
Ftemp [cur]. Fish = 0;
Time [I] [cur] ++;
Temp --;
}
}
Max =-1;
Cur = 0;
For (I = 0; I <n; I ++ ){
If (sum [I]> MAX ){
Max = sum [I];
Cur = I;
}
}
Cout <time [cur] [0] * 5;
For (I = 1; I <n; I ++)
Cout <"," <time [cur] [I] * 5;
Cout <Endl <"number of fish expected:" <sum [cur] <Endl;
}

}

 

Error code block:

Maxf = ftemp [0]. fish;
Cur = 0;
For (j = 1; j <= I; j ++ ){
If (maxf <ftemp [J]. Fish ){
Maxf = ftemp [J]. fish;
Cur = J;
}
}
If (ftemp [cur]. Fish <= 0 ){
Ftemp [cur]. Fish = 0;
Cur = 0;
}
Sum [I] + = ftemp [cur]. fish;
Ftemp [cur]. Fish-= ftemp [cur]. DESC;
Time [I] [cur] ++;

Temp --;