HDU1027 Ignatius and the Princess II 【next_permutation】【DFS】
Ignatius and the Princess II
you can work them out, I will release the Princess, or you will be my dinner, too." Ignatius says confidently, "OK, at last, I will save the Princess."
"Now I will show you the first problem." feng5166 says, "Given a sequence of number 1 to N, we define that 1,2,3...N-1,N is the smallest sequence among all the sequence which can be composed with number 1 to N(each number can be and should be use only once
in this problem). So it's easy to see the second smallest sequence is 1,2,3...N,N-1. Now I will give you two numbers, N and M. You should tell me the Mth smallest sequence which is composed with number 1 to N. It's easy, isn't is? Hahahahaha......"
Can you help Ignatius to solve this problem?
file.
6 4
11 8
1 2 3 5 6 4
1 2 3 4 5 6 7 9 8 11 10
题意:求数列1~n的第m个全排列。
题解:用STL里的next_permutation和DFS都可解,但时间复杂度就相去甚远了。
STL:耗时31ms
#include <cstdio>
#include <algorithm>
using namespace std; int arr[1002]; int main()
{
int n, m, i;
while(scanf("%d%d", &n, &m) != EOF){
for(i = 0; i < n; ++i)
arr[i] = i + 1;
while(--m) next_permutation(arr, arr + n);
for(i = 0; i < n; ++i)
if(i != n - 1) printf("%d ", arr[i]);
else printf("%d\n", arr[i]);
}
return 0;
}
DFS:耗时390ms
#include <stdio.h>
#include <string.h>
#define maxn 1002 int vis[maxn], n, m, arr[maxn], count, ok; void PRINT()
{
for(int i = 1; i <= n; ++i)
if(i != n) printf("%d ", arr[i]);
else printf("%d\n", arr[i]);
} void DFS(int k)
{
if(k > n) return;
for(int i = 1; i <= n; ++i){
if(!vis[i]){
vis[i] = 1; arr[k] = i;
if(k == n && m == ++count){
PRINT(); ok = 1; return;
}
DFS(k + 1); vis[i] = 0;
if(ok) return;
}
}
} int main()
{
while(scanf("%d%d", &n, &m) != EOF){
memset(vis, 0, sizeof(vis));
ok = count = 0; DFS(1);
}
return 0;
}
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