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ural 1142. Relations

2015-12-25 20:13 393 查看

1142. Relations

Time limit: 1.0 second
Memory limit: 64 MB

Background

Consider a specific set of comparable objects. Between two objects a and b, there exits one of the following three classified relations:

a = b
a < b
b < a

Because relation '=' is symmetric, it is not repeated above.

So, with 3 objects (a, b, c), there can exist 13 classified relations:

a = b = c a = b < c c < a = b a < b = c
b = c < a a = c < b b < a = c a < b < c
a < c < b b < a < c b < c < a c < a < b
c < b < a

Problem

Given N, determine the number of different classified relations between N objects.

Input

Includes many integers N (in the range from 2 to 10), each number on one line. Ends with −1.

Output

For each N of input, print the number of classified relations found, each number on one line.

Sample

input	output
2 3 -1	3 13

Tags: none (hide tags for unsolved problems)
Difficulty: 144

题意：计算N个数的大小比较关系的可能情况的总数。
分析：n个数，中间插入一些小于号。其余位置用等号。
等号相连的数的位置没有区别。
问题相当于问将n个数分为几个部分有多少种办法。
dp即可
dp[i][j]表示i个数，分为j组的方案数。
dp[i][j] += dp[i - 1][j] * j 表示第i个数有j中选择加入。
dp[i][j] += dp[i-1][j-1] * j 表示第i个数自立门户，然后这个新的部分插入其他j-1个已经确立大小关系的部分的方法有j种

/**
Create By yzx - stupidboy
*/
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define ft first
#define sd second
#define mk make_pair

inline int Getint()
{
int Ret = 0;
char Ch = ' ';
bool Flag = 0;
while(!(Ch >= '0' && Ch <= '9'))
{
if(Ch == '-') Flag ^= 1;
Ch = getchar();
}
while(Ch >= '0' && Ch <= '9')
{
Ret = Ret * 10 + Ch - '0';
Ch = getchar();
}
return Flag ? -Ret : Ret;
}

const int N = 15;
int n;
int dp

, ans
;

inline void Init()
{
dp[0][0] = 1;
for(int i = 1; i <= 10; i++)
for(int j = 1; j <= i; j++)
dp[i][j] = dp[i - 1][j] * j + dp[i - 1][j - 1] * j;
for(int i = 1; i <= 10; i++)
for(int j = 1; j <= i; j++)
ans[i] += dp[i][j];
}

inline void Solve();

inline void Input()
{
Init();
while(scanf("%d", &n) == 1)
{
if(n == -1) return;
Solve();
}
}

inline void Solve()
{
printf("%d\n", ans
);
}

int main()
{
freopen("a.in", "r", stdin);
Input();
//Solve();
return 0;
}

View Code

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