SRM 494 DIV.1 总结

250p：暴力。。。直接O(n^5)。。n<=50。

刚开始在想O(n^3)算法。。。搞了半天没搞出，最后还是用最暴力的方法写的，下次应该记得直接超暴力！

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string.h>

using namespace std;

class Painting {
public:
	int largestBrush(vector <string>);
};

int min (int a, int b)  {
	return a > b ? b : a;
}
int max (int a, int b)  {
	return a > b ? a : b;
}

bool vis[55][55];
int Painting::largestBrush(vector <string> p) {
	int i, j, k, l, o;
	int n = p.size();
	int m = p[0].size();
	int tot = min(n, m);
	for(i = tot;i >= 1 ;i--) {
		memset(vis, 0, sizeof(vis));
		for(j = 0;j < n; j ++) {
			for(k = 0;k < m; k++) {	
				if(j+i <= n && k+i <= m) {
					int ok = 1;
					for(l = j;l < j+i;l++) {
						for(o = k;o < k+i;o ++) {
							if(p[l][o] == 'W')
								ok = 0;
						}
					}
					if(ok) {
						for(l = j;l < j+i;l++) {
							for(o = k;o < k+i;o ++) {
								vis[l][o] = 1;
							}
						}
					}
				}
			}
		}
		int ok = 1;
		for(j = 0;j < n; j++) {
			for(k = 0;k < m; k++) {
				if(p[j][k] == 'B' && !vis[j][k])
					ok = 0;
			}
		}
		if(ok)
			return i;
	}
}


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题意：定义相邻两数严格单增单减交替的数列为交替数列，定义美值为交替数列中所有相邻两数差值绝对值的总和。

现在有最多50个数，每个数的值是是个随机整数，在区间low[i], high[i]，其中1<=low<=high<=1e5。如果这个数列的值组成的序列不是交替序列，则需要任意删掉一些使得其变成交替序列，求最后的序列的美值的期望是多少。

其实对于非交替数列，可以画图知道，无论怎么删除，总的美值大小还是等于所有相邻两数的差值绝对值总和，知道了这个，题目就顿时变得非常简单了。。。可以枚举前一个数的范围域，利用等差数列求和可得。

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <sstream>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <bitset>
#include <algorithm>
using namespace std;
#define clr(a, x) memset(a, x, sizeof(a))
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
template <class T> void checkmax(T &t, T x) { if (x > t) t = x; }
template <class T> void checkmin(T &t, T x) { if (x < t) t = x; }
template <class T> void _checkmax(T &t, T x) { if (t == -1 || x > t) t = x; }
template <class T> void _checkmin(T &t, T x) { if (t == -1 || x < t) t = x; }
typedef long long ll;
class AlternatingLane { 
    public: 
    double expectedBeauty(vector <int> low, vector <int> high);
    
// BEGIN CUT HERE
	public:
	void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
	private:
	template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
	void verify_case(int Case, const double &Expected, const double &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
	void test_case_0() { int Arr0[] = {1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {100}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 0.0; verify_case(0, Arg2, expectedBeauty(Arg0, Arg1)); }
	void test_case_1() { int Arr0[] = {1, 1, 1}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {2, 2, 2}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 1.0; verify_case(1, Arg2, expectedBeauty(Arg0, Arg1)); }
	void test_case_2() { int Arr0[] = {1, 3, 5, 7, 9}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {2, 4, 6, 8, 10}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 8.0; verify_case(2, Arg2, expectedBeauty(Arg0, Arg1)); }
	void test_case_3() { int Arr0[] = {4, 3, 3, 7}; vector <int> Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = {10, 7, 7, 7}; vector <int> Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); double Arg2 = 6.171428571428572; verify_case(3, Arg2, expectedBeauty(Arg0, Arg1)); }

// END CUT HERE
 
}; 

double cal(int n) {
	return (double)n*(n+1)/2;
}
double AlternatingLane::expectedBeauty(vector <int> l, vector <int> h) {
	int n = l.size();
	double ans = 0;
	int i, j, k;
	for(i = 0;i < n-1; i++) {
		double cur = 0;
		for(j = l[i];j <= h[i]; j++) {
			if(j < l[i+1])
				cur += cal(h[i+1]-j) - cal(l[i+1]-j-1);
			else if(j > h[i+1])
				cur += cal(j-l[i+1]) - cal(j-h[i+1]-1);
			else
				cur += cal(h[i+1]-j) + cal(j-l[i+1]);
		}
		cur /= (h[i]-l[i]+1);
		cur /= (h[i+1]-l[i+1]+1);
		ans += cur;
	}
	cout<<ans<<endl;
	return ans;
}

    // BEGIN CUT HERE 
int main() {
    AlternatingLane ___test; 
    ___test.run_test(-1); 
} 
    // END CUT HERE