[POJ3093 Margaritas on the River Walk]

[题目来源]：Greater New York 2006

[关键字]：背包

[题目大意]：求出使背包不能再放东西的方案数

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[分析]：一开始想求出背包容量为i时的方案数，但后来发现这样无法判断是否还有可以放入背包里的东西。看了题解后才明白要枚举。先将物品按体积升序排序，在从小到大枚举每一个最后剩余的不能装入背包的物品，此时i-1个物品必已装入背包，所以再对i+1件物品，sum-s[i-1]（部分和）容量的背包，进行01背包求解方案数，并对[max(s[i-1],sum-s[i]+1),m]区间的方案数累加就行了（看不懂就看代码，我就是一边看一边就理解了）。注意s[1]>sum的情况，和一些小细节。

[代码]：

View Code

 1 program Project1;
 2 var
 3   tc, n, m, i: longint;
 4   v: array[0..2000] of longint;
 5   s: array[0..2000] of longint;
 6   f: array[0..2000] of longint;
 7 
 8 procedure init;
 9 var
10   i, j, t: longint;
11 begin
12   read(n,m);
13   for i := 1 to n do read(v[i]);
14   for i := 1 to n-1 do
15     for j := i+1 to n do
16       if v[i] > v[j] then
17         begin
18           t := v[i];
19           v[i] := v[j];
20           v[j] := t;
21         end;
22   s[0] := 0;
23   for i := 1 to n do s[i] := s[i-1]+v[i];
24   //for i := 1 to n do write(v[i],' ');
25   //readln;
26 end;
27 
28 procedure work(tc: longint);
29 var
30   i, j, k, st, ans: longint;
31 begin
32   ans := 0;
33   for i := 1 to n do
34     begin
35       if s[i-1] > m then break;
36       fillchar(f,sizeof(f),0);
37       f[s[i-1]] := 1;
38       for j := i+1 to n do
39         for k := m downto s[i-1]+v[j] do
40           f[k] := f[k]+f[k-v[j]];
41       if s[i-1] < m-v[i]+1 then st := m-v[i]+1 else st := s[i-1];
42       for j := st to m do
43         inc(ans,f[j]);
44     end;
45   if s[1] > m then ans := 0;
46   writeln(tc,' ',ans);
47 end;
48 
49 begin
50   //assign(output,'d:\1.out');rewrite(output);
51   readln(tc);
52   for i := 1 to tc do
53     begin
54       init;
55       work(i);
56     end;
57   //close(output);
58 end.