【POJ 3187 Backward Digit Sums】 + DFS + 排列组合

最新推荐文章于 2018-09-07 15:58:00 发布

原创最新推荐文章于 2018-09-07 15:58:00 发布 · 381 阅读

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#dfs #poj #numbers

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本文介绍了一款名为BackwardDigitSums的数学游戏，玩家需从最终总和逆推原始数字序列。通过递归深度优先搜索算法，实现寻找满足条件的最小字典序数字序列，并提供完整的AC代码。

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Backward Digit Sums
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7041 Accepted: 4083

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this:

3   1   2   4
  4   3   6
    7   9
     16

Behind FJ’s back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ’s mental arithmetic capabilities.

Write a program to help FJ play the game and keep up with the cows.
Input

Line 1: Two space-separated integers: N and the final sum.
Output

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.
Sample Input

4 16
Sample Output

3 1 2 4
Hint

Explanation of the sample:

There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

顶层的数 = C(N - 1,0) * a[1] + C(N - 1,1) * a[2]+ …+ C(N - 1,N - 1) * a[N]~~ 字典序最小，对底层数dfs？

AC代码：

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
int dp[15][15],a[15],M,N,vis[15],ok,b[15];
void dfs(int n,int cut){
    if(n == N && cut == M && ok){
        ok = 0;
        for(int i = 0; i < N; i++) printf("%d ",a[i]);
        printf("\n");
    }
    if(n == N || cut > M || !ok) return ;
    for(int i = 1; i <= N; i++)
      if(!vis[i]){
           vis[i] = 1;
           a[n] = i;
           dfs(n + 1,cut + dp[N - 1][n] * i);
           vis[i] = 0;
      }
    return ;
}
int main()
{
    dp[1][1] = 1;
    for(int i = 0; i <= 12; i++)
        dp[i][0] = 1;
    for(int i = 2; i <= 12; i++)
        for(int j = 1; j <= i; j++)
            dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
   while(scanf("%d %d",&N,&M) != EOF){
        memset(vis,0,sizeof(vis));
        ok = 1;
        dfs(0,0);
   }
    return 0;
}