Codeforces-258D:Little Elephant and Broken Sorting(概率DP)
2017-10-16 17:31
537 查看
D. Little Elephant and Broken Sorting
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
The Little Elephant loves permutations of integers from 1 to n very
much. But most of all he loves sorting them. To sort a permutation, the Little Elephant repeatedly swaps some elements. As a result, he must receive a permutation 1, 2, 3, ..., n.
This time the Little Elephant has permutation p1, p2, ..., pn.
Its sorting program needs to make exactly m moves, during the i-th
move it swaps elements that are at that moment located at the ai-th
and the bi-th
positions. But the Little Elephant's sorting program happened to break down and now on every step it can equiprobably either do nothing or swap the required elements.
Now the Little Elephant doesn't even hope that the program will sort the permutation, but he still wonders: if he runs the program and gets some permutation, how much will the result of sorting resemble the sorted one? For that help the Little Elephant find
the mathematical expectation of the number of permutation inversions after all moves of the program are completed.
We'll call a pair of integers i, j (1 ≤ i < j ≤ n) an inversion in
permutatuon p1, p2, ..., pn,
if the following inequality holds: pi > pj.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1000, n > 1) —
the permutation size and the number of moves. The second line contains n distinct integers, not exceeding n —
the initial permutation. Next m lines each contain two integers: the i-th
line contains integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) —
the positions of elements that were changed during the i-th move.
Output
In the only line print a single real number — the answer to the problem. The answer will be considered correct if its relative or absolute error does not exceed 10 - 6.
Examples
input
output
input
output
思路:f[i][j]表示第i个数比第j个数大的概率。
#include<bits/stdc++.h>
using namespace std;
const int MAX=1e3+10;
double f[MAX][MAX];
int n,m,A[MAX];
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)scanf("%d",&A[i]);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)f[i][j]=(A[i]>A[j]);
}
while(m--)
{
int x,y;
scanf("%d%d",&x,&y);
for(int i=1;i<=n;i++)
{
if(i==x||i==y)continue;
f[i][x]=f[i][y]=(f[i][x]+f[i][y])/2;
f[x][i]=f[y][i]=(f[x][i]+f[y][i])/2;
}
f[x][y]=f[y][x]=0.5;
}
double ans=0;
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)ans+=f[i][j];
}
printf("%0.7lf\n",ans);
return 0;
}
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
The Little Elephant loves permutations of integers from 1 to n very
much. But most of all he loves sorting them. To sort a permutation, the Little Elephant repeatedly swaps some elements. As a result, he must receive a permutation 1, 2, 3, ..., n.
This time the Little Elephant has permutation p1, p2, ..., pn.
Its sorting program needs to make exactly m moves, during the i-th
move it swaps elements that are at that moment located at the ai-th
and the bi-th
positions. But the Little Elephant's sorting program happened to break down and now on every step it can equiprobably either do nothing or swap the required elements.
Now the Little Elephant doesn't even hope that the program will sort the permutation, but he still wonders: if he runs the program and gets some permutation, how much will the result of sorting resemble the sorted one? For that help the Little Elephant find
the mathematical expectation of the number of permutation inversions after all moves of the program are completed.
We'll call a pair of integers i, j (1 ≤ i < j ≤ n) an inversion in
permutatuon p1, p2, ..., pn,
if the following inequality holds: pi > pj.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 1000, n > 1) —
the permutation size and the number of moves. The second line contains n distinct integers, not exceeding n —
the initial permutation. Next m lines each contain two integers: the i-th
line contains integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) —
the positions of elements that were changed during the i-th move.
Output
In the only line print a single real number — the answer to the problem. The answer will be considered correct if its relative or absolute error does not exceed 10 - 6.
Examples
input
2 1 1 2 1 2
output
0.500000000
input
4 3 1 3 2 4 1 2 2 3 1 4
output
3.000000000
思路:f[i][j]表示第i个数比第j个数大的概率。
#include<bits/stdc++.h>
using namespace std;
const int MAX=1e3+10;
double f[MAX][MAX];
int n,m,A[MAX];
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)scanf("%d",&A[i]);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)f[i][j]=(A[i]>A[j]);
}
while(m--)
{
int x,y;
scanf("%d%d",&x,&y);
for(int i=1;i<=n;i++)
{
if(i==x||i==y)continue;
f[i][x]=f[i][y]=(f[i][x]+f[i][y])/2;
f[x][i]=f[y][i]=(f[x][i]+f[y][i])/2;
}
f[x][y]=f[y][x]=0.5;
}
double ans=0;
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)ans+=f[i][j];
}
printf("%0.7lf\n",ans);
return 0;
}
相关文章推荐
- CodeForces 258 B.Little Elephant and Elections(数位DP+dfs)
- Codeforces--Little Elephant and Sorting
- CodeForces - 205B - Little Elephant and Sorting
- Codeforces 221 B. Little Elephant and Numbers
- codeforces 258-E. Little Elephant and Tree 线段树,离线
- <codeforces>Little Elephant and Sorting
- CodeForces 220A Little Elephant and Problem
- Codeforces 221 C. Little Elephant and Problem
- CodeForces 204B - Little Elephant and Cards
- CodeForces 220B Little Elephant and Array
- Codeforces 221 D. Little Elephant and Array
- Codeforces Round #129 (Div. 2) B. Little Elephant and Sorting
- codeforces 221D Little Elephant and Array
- Codeforces-258C:Little Elephant and LCM(数论)
- 线段树 CodeForces 220B - Little Elephant and Array
- CodeForces 220B Little Elephant and Array 莫队算法
- Codeforces 518D Ilya and Escalator【概率dp】
- SRM 595 D2 L3:LittleElephantAndXor, dp
- CodeForces 204A - Little Elephant and Interval 算答案..易错..暴力打表对拍
- CodeForces 167B - Wizards and Huge Prize 期望概率dp