Codeforces 258D Little Elephant and Broken Sorting （看题解） 概率dp

时间：2022-10-01 08:02:37

Little Elephant and Broken Sorting

怎么感觉这个状态好难想到啊。。

dp[ i ][ j ]表示第 i 个数字比第 j 个数字大的概率。转移好像比较显然。

#include<bits/stdc++.h>

#define LL long long

#define LD long double

#define ull unsigned long long

#define fi first

#define se second

#define mk make_pair

#define PLL pair<LL, LL>

#define PLI pair<LL, int>

#define PII pair<int, int>

#define SZ(x) ((int)x.size())

#define ALL(x) (x).begin(), (x).end()

#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N =  + ;

const int inf = 0x3f3f3f3f;

const LL INF = 0x3f3f3f3f3f3f3f3f;

const int mod = 1e9 + ;

const double eps = 1e-;

const double PI = acos(-);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}

template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < ) a += mod;}

template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}

template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m, a[N];

double dp[N][N];

int main() {

    scanf("%d%d", &n, &m);

    for(int i = ; i <= n; i++) scanf("%d", &a[i]);

    for(int i = ; i <= n; i++)

        for(int j = ; j <= n; j++)

            if(a[i] > a[j]) dp[i][j] = ;

    while(m--) {

        int a, b; scanf("%d%d", &a, &b);

        for(int i = ; i <= n; i++) {

            if(i == a || i == b) continue;

            dp[i][a] = dp[i][b] = (dp[i][a] + dp[i][b]) / ;

            dp[a][i] = dp[b][i] = (dp[a][i] + dp[b][i]) / ;

        }

        dp[a][b] = dp[b][a] = 0.5;

    }

    double ans = ;

    for(int i = ; i <= n; i++)

        for(int j = i + ; j <= n; j++)

            ans += dp[i][j];

    printf("%.12f\n", ans);

    return ;

}

/*

*/

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