CodeForces 351 B.Jeff and Furik(概率DP)
2017-04-11 19:21
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Description
给一个1~n的排列p[i],Jeff先手可以交换任意两个相邻元素,而Furik会有0.5的几率把任意满足p[i] < p[i+1]的p[i]和p[i+1]交换,有0.5的几率把任意满足p[i] > p[i+1]的p[i]和p[i+1]交换,问将整个序列变成升序所需的最小期望步数
Input
第一行一整数n表示序列长度,之后一个1~n的排列p[i] (1<=n<=3000)
Output
输出把整个序列变成升序所需的最小期望步数
Sample Input
2
1 2
Sample Output
0.000000
Solution
Jeff一次操作必然减少一个逆序对,而Furik一次操作0.5几率减少一个逆序对,0.5几率增加一个逆序对,设dp[i]为序列逆序对数为i时将序列变成升序所需的最小期望步数,那么有dp[0]=0,dp[1]=1,dp[i]=0.5*dp[i-2]+0.5*dp[i]+2,即dp[i]=dp[i-2]+4,故dp数组分奇偶后就是两个等差数列,dp[i]=2*i(i为偶数),dp[i]=2*i-1(i为奇数),逆序对直接n^2暴力即可
Code
给一个1~n的排列p[i],Jeff先手可以交换任意两个相邻元素,而Furik会有0.5的几率把任意满足p[i] < p[i+1]的p[i]和p[i+1]交换,有0.5的几率把任意满足p[i] > p[i+1]的p[i]和p[i+1]交换,问将整个序列变成升序所需的最小期望步数
Input
第一行一整数n表示序列长度,之后一个1~n的排列p[i] (1<=n<=3000)
Output
输出把整个序列变成升序所需的最小期望步数
Sample Input
2
1 2
Sample Output
0.000000
Solution
Jeff一次操作必然减少一个逆序对,而Furik一次操作0.5几率减少一个逆序对,0.5几率增加一个逆序对,设dp[i]为序列逆序对数为i时将序列变成升序所需的最小期望步数,那么有dp[0]=0,dp[1]=1,dp[i]=0.5*dp[i-2]+0.5*dp[i]+2,即dp[i]=dp[i-2]+4,故dp数组分奇偶后就是两个等差数列,dp[i]=2*i(i为偶数),dp[i]=2*i-1(i为奇数),逆序对直接n^2暴力即可
Code
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<queue> #include<map> #include<set> #include<ctime> using namespace std; typedef long long ll; #define INF 0x3f3f3f3f #define maxn 3333 int n,a[maxn]; int main() { while(~scanf("%d",&n)) { for(int i=1;i<=n;i++)scanf("%d",&a[i]); int ans=0; for(int i=1;i<=n;i++) for(int j=1;j<i;j++) if(a[j]>a[i])ans++; if(ans%2==0)ans*=2; else ans=2*ans-1; printf("%d\n",ans); } return 0; }
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