全排列next_permutation的用法 HDU 1027 PKU 3187 1146 1731

全排列next_permutation HDU 1027 PKU 3187 1146 1731

(2009-11-22 11:42:40)
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全排列

杂谈

分类:STL

这个函数可以计算一组数据的全排列

 

假设数列 d1,d2,d3,d4……

范围由[first,last)标记,调用next_permutation使数列逐次增大,这个递增过程按照字典序。

 

若当前调用排列到达最大字典序,比如dcba,就返回false,同时重新设置该排列为最小字典序。

返回为true表示生成下一排列成功。

 

另外,库中另一函数prev_permutation与next_permutation相反,由原排列得到字典序中上一次最近排列。

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几道题目:

 

1.HDU 1027

 求第M个排列,写错一次,数组开头num+1没+1全排列next_permutation <wbr>HDU <wbr>1027 <wbr>PKU <wbr>3187 <wbr>1146 <wbr>1731

代码:

#include
#include
using namespace std;
int main()
{
 int n,m,i;
 int num[1001];
 for(i=1;i<=1000;i++)
  num[i]=i;
 while(scanf("%d%d",&n,&m)!=EOF)
 {
  for(i=0;i   next_permutation(num+1,num+n+1);
  for(i=1;i<=n;i++)
  {
   if(i>1) printf(" ");
   printf("%d",num[i]);
  }
  printf("/n");
  sort(num+1,num+n+1);
 }
 return 0;
}

 

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2.PKU 3187

Backward Digit Sums
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 1653   Accepted: 950

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.

 

 

看这题目的风格似乎是USACO上的题?还没做到,预先过了一题,嘿嘿

 

生成N的全排列,然后一次计算s[0],与sum相等即为答案!

 

调试一次,把s[]和m的初始化放到了while(1)的外面,变成死循环

代码:

#include
#include
#include
using namespace std;
int main()
{
 int n,sum,i,j,m;
 int num[11],s[11];
 for(i=0;i<10;i++)
  num[i]=i+1;
 while(scanf("%d%d",&n,&sum)!=EOF)
 {
  while(1)
  {
   for(i=0;i    s[i]=num[i];
   m=n-1;
   while(m!=0)
   {
    for(j=0;j     s[j]+=s[j+1];
    m--;
   }
   if(s[0]==sum)
 break;
   next_permutation(num,num+n);
  }
  for(i=0;i  {
   if(i>0) printf(" ");
   printf("%d",num[i]);
  }
  printf("/n");
 }
 return 0;
}

**********************************************************************************************

PKU 1146

ID Codes
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 2657   Accepted: 1645

简单应用

代码:

#include
#include
using namespace std;
int main()
{
 int len;
 char s[51];
 while(scanf("%s",s)!=EOF)
 {
  if(s[0]=='#')
   break;
  len=strlen(s);
  if(next_permutation(s,s+len)) 
   printf("%s/n",s);
  else
   printf("No Successor/n");
 }
 return 0;
}

 

**********************************************************************************************

PKU 1731

Orders
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 4751   Accepted: 3060

 

输出全排列,一开始打算用prev_permutation回到字典序最小的那个,发现循环停止不了

是不是prev_permutation的返回值跟next的不同呢?解决后再补上

于是改用sort排序,也是一题简单应用

 

代码:

#include
#include
#include
using namespace std;
bool cmp(char x,char y)
{
 return x}
int main()
{
 int len;
 char s[201];
 while(scanf("%s",s)!=EOF)
 {
  len=strlen(s);
  sort(s,s+len,cmp);
  printf("%s/n",s);
  while(next_permutation(s,s+len))
   printf("%s/n",s);
 }
 return 0;
}

 

转载自:http://blog.sina.com.cn/s/blog_62a14e150100fti9.html

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