动态规划练习题(1)

1、0/1背包(package.pas)

AYYZOJ p1472

 1 program p1472;
 2 const
 3   maxm=200; maxn=30;
 4 var
 5   m,n,i,j:integer;
 6   c,w:array[1..maxn] of integer;
 7   f:array[0..maxn,0..maxm] of integer;
 8 function max(x,y:integer):integer;
 9 begin
10   if x>y then max:=x else max:=y;
11 end;
12 begin
13   readln(m,n);
14   for i:=1 to n do readln(w[i],c[i]);
15   for i:=1 to n do
16    for j:=1 to m do
17     begin
18       if j>=w[i] then f[i,j]:=max(f[i-1,j-w[i]]+c[i],f[i-1,j])
19         else f[i,j]:=f[i-1,j];
20     end;
21  writeln(f[n,m]);
22 end.

参考程序

2、采药(medic.pas)

AYYZOJ p1052

COGS p68

 1 const
 2   maxm=1000; maxn=100;
 3 var
 4   f:array[0..maxn,0..maxm] of integer;
 5   w,c:array[1..maxn] of integer;
 6   m,n,i,j:integer;
 7
 8 function max(x,y:integer):integer;
 9 begin
10   if x>y then max:=x else max:=y
11 end;
12
13 begin
14 assign(input,‘medic.in‘);
15 reset(input);
16 assign(output,‘medic.out‘);
17 rewrite(output);
18   readln(m,n);
19   for i:=1 to n do readln(w[i],c[i]);
20   fillchar(f,sizeof(f),0);
21    for i:=1 to n do
22     for j:=1 to m do
23      if j>=w[i] then f[i,j]:=max(f[i-1,j-w[i]]+c[i],f[i-1,j])
24                else f[i,j]:=f[i-1,j];
25   writeln(f[n,m]);
26 close(input);
27 close(output);
28 end.

参考程序

3、开心的金明(happy.pas)

AYYZOJ p1059

COGS p71

 1 var
 2   f:array[0..30000] of longint;
 3   w,c:array[1..30000] of longint;
 4   i,j,m,n,ans:longint;
 5 begin
 6 assign(input,‘happy.in‘);
 7 reset(input);
 8 assign(output,‘happy.out‘);
 9 rewrite(output);
10   readln(m,n);
11   for i:=1 to n do readln(w[i],c[i]);
12   fillchar(f,sizeof(f),0);
13    for i:=1 to n do
14     for j:=m downto w[i] do
15      if f[j-w[i]]+c[i]*w[i]>f[j] then f[j]:=f[j-w[i]]+c[i]*w[i]
16       else f[j]:=f[j]+c[j]*w[j];
17    writeln(f[m]);
18 close(input);
19 close(output);
20 end.

参考程序1

 1 var
 2   n,m,i,j:integer;
 3   v,p:array[1..25] of longint;
 4   dp:array[0..30000] of longint;
 5   function max(x,y:longint):longint;
 6   begin if x>y then max:=x else max:=y; end;
 7 begin
 8   assign(input,‘happy.in‘); assign(output,‘happy.out‘);
 9   reset(input); rewrite(output);
10   readln(n,m);
11   for i:=1 to m do
12     begin
13       readln(v[i],p[i]);
14       p[i]:=p[i]*v[i];
15     end;
16   for i:=1 to m do
17     for j:=n downto v[i] do
18       dp[j]:=max(dp[j],dp[j-v[i]]+p[i]);
19   writeln(dp[n]);
20   close(input); close(output);
21 end.

参考程序2

4、装箱问题(box.pas)

AYYZOJ p1021

COGS p1089

我的思路：使剩余空间最小，则可以设每件物品的价值c[i]=每件物品的体积w[i] 。转化为0/1背包问题，最后输出V-dp[v]即可

 1 var
 2   v,n,i,j:longint;
 3   w,c:array[1..30] of longint;
 4   dp:array[0..20000] of longint;
 5 function max(x,y:longint):longint;
 6 begin
 7   if x>y then max:=x else max:=y;
 8 end;
 9 begin
10 assign(input,‘npack.in‘);
11 reset(input);
12 assign(output,‘npack.out‘);
13 rewrite(output);
14   readln(v);
15   readln(n);
16   for i:=1 to n do begin readln(w[i]); c[i]:=w[i]; end;
17    for i:=1 to n do
18     for j:=v downto w[i] do
19      dp[j]:=max(dp[j-w[i]]+c[i],dp[j]);
20    writeln(v-dp[v]);
21 close(input);
22 close(output);
23 end.

我的程序

另一种思路：{f[j]是布尔数组，表示前i件物品正好装满容积为j的箱子,最后从后往前扫描布尔数组f,找到最大值}

 1 var
 2  f:array [0..20000] of boolean;
 3  box:array [0..30] of integer;
 4  n,v,m,i,j:longint;
 5 begin
 6  readln(v);
 7  readln(n);
 8  for i:=1 to n do read(box[i]);
 9  for i:=0 to v do f[i]:=false;
10  f[0]:=true;
11  for i:=1 to n do
12    for j:=v downto box[i] do
13      if f[j-box[i]] then f[j]:=true;
14  for i:=v downto 0 do if f[i]=true then
15    begin
16       writeln(v-i);
17       halt;
18    end;
19 end.

老师给的程序

5、竞赛真理（truth.pas/c/cpp）

AYYZOJ p1493

 1 var
 2   n,t,i,j:longint;
 3   f:array[0..1080000] of longint;
 4   w1,t1,w2,t2:array[0..30] of longint;
 5 function max(a,b:longint):longint;
 6 begin
 7   if a>b then max:=a
 8    else max:=b;
 9 end;
10 begin
11   fillchar(f,sizeof(f),0);
12   readln(n,t);
13   for i:=1 to n do
14    readln(w1[i],t1[i],w2[i],t2[i]);
15   for i:=1 to n do
16    for j:=t downto 1 do
17    begin
18      if j>=t1[i] then
19       f[j]:=max(f[j-t1[i]]+w1[i],f[j]);
20      if j>=t2[i] then
21       f[j]:=max(f[j-t2[i]]+w2[i],f[j]);
22    end;
23   writeln(f[t]);
24 end.

我的程序

 1 var a,b:array[1..30,1..2] of longint;
 2     f:array[0..1080000] of longint;
 3     i,j,n,time:longint;
 4 begin
 5   readln(n,time);
 6   for i:=1 to n do
 7     read(a[i,1],b[i,1],a[i,2],b[i,2]);
 8   for i:=1 to n do
 9     for j:=time downto 1 do begin
10       if j>=b[i,1] then
11         if f[j]<f[j-b[i,1]]+a[i,1] then f[j]:=f[j-b[i,1]]+a[i,1];
12       if j>=b[i,2] then
13         if f[j]<f[j-b[i,2]]+a[i,2] then f[j]:=f[j-b[i,2]]+a[i,2];
14     end;
15   writeln(f[time]);
16 end.

老师给的程序