[hdu5216]排序

题意：给定两个长度为M的数组a,b，对于一个1-M的排列，不妨设为P，如果对任意0<=i<M，都有a[i] <= b[P[i]]，那么称为一种合法情况，对于一种合法情况，对所有0<=i<M,在n个长度为1的线段上的区间[a[i],b[p[i]]]涂上颜色，计X=没有涂颜色的最大连续长度，求x在所有合法情况中的期望。

思路：这个题想到了就是大水题了，可惜比赛的时候题目都没看。由于全排列P的存在，使得a数组可以对应b数组的任意一种“比较方式”，于是存在合法情况等价于存在一种b数组的全排列使得a[i]<=b[i]恒成立，由于全排列的任意性，不妨将a数组，b数组分别排序，如果对任意i，a[i]<=b[i]恒成立，那么合法情况存在。然后合法情况存在的基础上，考虑重新排列一下b数组，以得到其它的合法情况。在重排列过程中注意到，无论怎么重排，只要是合法情况，最后线段的涂色情况是一样的！于是对每一种合法情况，概率一样，X一样，所以期望等于任意一种合法情况的X。由于根据所有的i，把区间[a[i], b[i]]的线段涂上颜色，没涂颜色的连续段只可能出现在[b[i]+1,a[i+1]-1]（至于为什么，画个图就清楚了，相当于左右边界分别递增的线段去覆盖），用这个去更新答案。

#pragma comment(linker, "/STACK:102400000,102400000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>
#include <vector>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 1e5 + 7;
const int md = 10007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

int a[60], b[60];

int main() {
    //freopen("in.txt", "r", stdin);
    int T, n, m;
    cin >> T;
    while (T --) {
        cin >> n >> m;
        rep_up0(i, m) {
            sint(a[i]);
        }
        rep_up0(i, m) {
            sint(b[i]);
        }
        sort(a, a + m);
        sort(b, b + m);
        bool ok = true;
        rep_up0(i, m) {
            if (a[i] > b[i]) {
                ok = false;
                break;
            }
        }
        if (!ok) {
            puts("Stupid BrotherK!");
            continue;
        }
        int ans = a[0] - 1;
        rep_up0(i, m - 1) {
            max_update(ans, a[i + 1] - b[i] - 1);
        }
        max_update(ans, n - b[m - 1]);
        printf("%d.000000
", ans);
    }
    return 0;
}