66 lines
1.7 KiB
Plaintext
66 lines
1.7 KiB
Plaintext
Program Fishing3; {堆结构、贪心}
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Type
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Data = Record
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fish, lake: integer; //堆中结点的信息
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End;
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Var
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heap: Array[1..100] Of Data;
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t, f, d: Array[1..100] Of integer;
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ans, i, j, k, m, Max, n, t1, Time: Longint;
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Procedure Sift(i: Longint); //堆操作，本题只需要向下筛
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Var
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a: Data;
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j: Longint;
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Begin
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a := heap[i];
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j := i shl 1;
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While (j <= k) Do Begin
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If (j < k) and (heap[j].fish < heap[j + 1].fish) Then Inc(j);
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If a.fish < heap[j].fish Then Begin
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heap[i] := heap[j];
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i := j;
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j := i shl 1;
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End
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Else Break;
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End;
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heap[i] := a;
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End;
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Begin
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assign(input,'fishing.in');
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assign(output,'fishing.out');
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reset(input);
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rewrite(output);
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Readln(n);
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For i := 1 To n Do Read(f[i]);readln;
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For i := 1 To n Do Read(d[i]);readln;
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For i := 1 To n-1 Do Read(t[i]);t[n]:=0;readln;
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Readln(m);
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t1 := 0;
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Max := 0;
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For k := 1 To n Do Begin //枚举最远走到的池塘的编号
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Time := m - t1; //计算剩余时间
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ans := 0;
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For i := 1 To k Do Begin //收集能够钓鱼的池塘的资料
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heap[i].fish := f[i];
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heap[i].lake := i;
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End;
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For i := 1 To k shr 1 Do Sift(i); //堆的初始化
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While (Time > 0) and (heap[1].fish > 0) Do
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Begin
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Inc(ans, heap[1].fish); //贪心选取鱼最多的池塘
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Dec(heap[1].fish, d[heap[1].lake]); //修改鱼的数量
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Sift(1); //堆维护
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Dec(Time); //剩余时间变少
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End;
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If ans > Max Then Max := ans; //刷新最优解
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Inc(t1, t[k]); //累计走路需要的时间
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End;
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Writeln(Max);
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close(input);
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close(output)
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End.
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