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HDU 1028 Ignatius and the Princess III（传说中的母函数求方案数）

2013-05-27 17:59 453 查看
很有成就感的说。。现在再看简直觉得难以想象，当时居然能用bfs做。。。虽然TLE了。。

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9048    Accepted Submission(s): 6366


[align=left]Problem Description[/align]
"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:

  N=a[1]+a[2]+a[3]+...+a[m];

  a[i]>0,1<=m<=N;

My question is how many different equations you can find for a given N.

For example, assume N is 4, we can find:

  4 = 4;

  4 = 3 + 1;

  4 = 2 + 2;

  4 = 2 + 1 + 1;

  4 = 1 + 1 + 1 + 1;

so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

 

[align=left]Input[/align]
The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

 

[align=left]Output[/align]
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

 

[align=left]Sample Input[/align]

4 10 20

 

[align=left]Sample Output[/align]

5 42 627

 

[align=left]Author[/align]
Ignatius.L

Statistic | Submit | Discuss | Note
呵呵，用BFS和set，multiset，直接爆表了。。思路和那个第一次见很难理解的蛇出洞题很像，但这里用的是multiset，来保存访问信息。爆表爆表。。

[align=left]#include<iostream>[/align]
[align=left]#include<queue>[/align]
[align=left]#include<set>[/align]
[align=left]using namespace std;[/align]
#define MAXN
120
[align=left]int n,CNT;             //输入的规模n和结果CNT[/align]
set<multiset< char>
> S[MAXN+1];  //存储访问信息
[align=left]struct node[/align]
[align=left]{[/align]
[align=left]                 int x;                                                        //当前位置[/align]
  
4000
              multiset< char>
s; //节点自身路径
[align=left]}init,now,nextt;                                                     [/align]
[align=left]queue<node> Q;[/align]
[align=left]int BFS()[/align]
[align=left]{[/align]
                 for(int j=0;j<MAXN+1;++j)
S[j].clear();
[align=left]                CNT = 0;[/align]
                 for(int k=1;k<=n;++k)
                            //先放n个初始节点进去
[align=left]                {[/align]
[align=left]                                node nod;[/align]
[align=left]                                nod.x = k;[/align]
[align=left]                                nod.s.insert(k);[/align]
[align=left]                                Q.push(nod);[/align]
[align=left]                                S[k].insert(nod.s);[/align]
[align=left]                }[/align]
[align=left]                init.x = 0;init.s.clear();[/align]
[align=left]                Q.push(init);[/align]
[align=left]                 while(!Q.empty())[/align]
[align=left]                {[/align]
[align=left]                                now = Q.front();Q.pop();[/align]
                                 if(now.x==n)
{CNT++;continue ;}
[align=left]                                 for(int i=1;i<=n;++i)[/align]
[align=left]                                {[/align]
[align=left]                                                nextt.x = now.x+i;[/align]
[align=left]                                                nextt.s = now.s;[/align]
[align=left]                                                nextt.s.insert(i);    [/align]
[align=left]                                                 if(nextt.x<=n&&S[nextt.x].count(nextt.s)==0)                 [/align]
[align=left]                                                {              [/align]
[align=left]                                                                Q.push(nextt);                                                      [/align]
[align=left]                                                                S[nextt.x].insert(nextt.s);     [/align]
[align=left]                                                }[/align]
[align=left]                                }[/align]
[align=left]                }[/align]
[align=left]                 return CNT;[/align]
[align=left]}[/align]
[align=left]int main()[/align]
[align=left]{[/align]
[align=left]                 while(cin>>n)[/align]
[align=left]                                cout<<BFS()<<endl;[/align]
[align=left]                 return 0;[/align]
[align=left]}[/align]

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