fzuoj 2149 Reverse Game

来源：互联网发布：课程表管理系统源码编辑：程序博客网时间：2024/06/05 23:50
题目戳这里
思路：矩阵乘法超时了。打了个表发现很快就可以收敛了，于是暴力迭代直到误差小于1e-6（这题从头到尾都是乱来）
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int dim;struct matrix{    double a[105][105];    void init() {        for(int i = 0; i <= dim; i ++)            for(int j = 0; j <= dim; j ++)                a[i][j] = 0;    }        matrix operator * (const matrix &p) {        matrix c;        c.init();        for(int i = 1; i <= dim; i ++)            for(int j = 1; j <= dim; j ++)                for(int k = 1; k <= dim; k ++)                    c.a[i][j] += a[i][k] * p.a[k][j];        return c;    }};matrix pow(int n,matrix t){    matrix b;    b.init();    for(int i = 1; i <= dim; i ++) b.a[i][i] = 1;    while(n) {        if(n & 1) b = b * t;        t = t * t;        n /= 2;    }    return b;}bool cmp(int x,int y){    return x > y;}double dp[5010][110];void solve(){    int n,m;    double a[110];    double b[110];    scanf("%d%d",&n,&m);    for(int i = 1; i <= n; i ++)        scanf("%lf",&a[i]);    sort(a + 1,a + n + 1,cmp);    int lt = (n + 1) / 2,rt = lt + 1;    for(int i = 1; i <= n; i ++)        if(i & 1) b[lt --] = a[i];        else b[rt ++] = a[i];    if(n <= 2) {        double ans = 0;        for(int i = 1; i <= n; i ++)            ans += a[i];        printf("%.6lf\n",ans);        return;    }    matrix c;    dim = n;    c.init();    //for(int i = 1; i <= n; i ++) cout << b[i] << " ";    //cout << endl;    double p = 0;    double pp[110];    memset(pp,0,sizeof(pp));    for(int i = 1; i < n; i ++)        for(int j = i + 1; j <= n; j ++)            for(int k = i; k <= j; k ++)                pp[k] ++;    p = n * (n - 1) / 2;     //for(int i = 1; i <= n; i ++)        //cout << pp[i] << " ";    //cout << endl;    for(int i = 1; i < n; i ++) {        for(int j = i + 1; j <= n; j ++) {            for(int k = i; k <= j; k ++) {                c.a[j - (k - i)][k] += 1.0 / p;                //cout << j - (k - i) << " " << k << " " << 1.0 / p << endl;            }        }    }    for(int i = 1; i <= n; i ++) c.a[i][i] += (p - pp[i]) / p;    //for(int i = 1; i  <= n; i ++) {        //for(int j = 1; j <= n; j ++)            //cout << c.a[i][j] << " ";        //cout << endl;    //}    if(m) {        for(int i = 1; i <= n; i ++) dp[0][i] = b[i];        for(int i = 1; i <= m; i ++) {            bool flag = true;            for(int j = 1; j <= n; j ++)                dp[i][j] = 0;            for(int j = 1; j <= n; j ++)                for(int k = 1; k <= n; k ++)                    dp[i][j] += c.a[j][k] * dp[i - 1][k];            for(int j = 1; j <= n && flag; j ++)                    if(fabs(dp[i][j] - dp[i - 1][j]) > 1e-6) {                        flag = false;                        break;                    }            if(flag || i == m) {                for(int j = 1; j <= n; j ++) a[j] = dp[i][j];                break;            }        }        /*        c = pow(m,c);        for(int i = 1; i <= n; i ++) {            a[i] = 0;            for(int j = 1; j <= n; j ++)                a[i] += c.a[i][j] * b[j];        }*/    }    else {        for(int i = 1; i <= n; i ++) a[i] = b[i];    }    //cout << endl;    //for(int i = 1; i  <= n; i ++) {        //for(int j = 1; j <= n; j ++)            //cout << c.a[i][j] << " ";        //cout << endl;    //}        //for(int i = 1; i <= n; i ++) cout << a[i] << " ";    //cout << endl;    double ans = 0;    a[0] = 0;    for(int i = 2; i <= n; i ++) a[i] += a[i - 1];    for(int i = 1; i <= n; i ++)        for(int j = i + 1; j <= n; j ++)            ans += a[j] - a[i - 1];    p = n * (n - 1) / 2;    ans /= p;    printf("%.6lf\n",ans);}int main(){    //freopen("1.txt","w",stdout);    int t;    scanf("%d",&t);    while(t --) solve();    return 0;}
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