Codeforces Round #329 (Div. 2)

推迟了15分钟开始，B卡住不会，最后弃疗，rating只涨一分。。。

 

水(暴力枚举) A - 2Char

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 21:33:17
* File Name     :A.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
char s[110][1010];
int len[110];
bool ok[110];
vector<char> vec[110];

int main(void)    {
	int n;	scanf ("%d", &n);
	for (int i=1; i<=n; ++i)	{
		scanf ("%s", s[i] + 1);
		len[i] = strlen (s[i] + 1);
	}
	memset (ok, true, sizeof (ok));
	int vis[30];
	for (int i=1; i<=n; ++i)	{
		memset (vis, 0, sizeof (vis));
		int tot = 0;
		for (int j=1; j<=len[i]; ++j)	{
			if (!vis[s[i][j]-'a'])	{
                vec[i].push_back (s[i][j]);
				vis[s[i][j]-'a'] = 1;	tot++;
			}
			else	continue;
			if (tot > 2)	{
				ok[i] = false;	break;
			}
		}
	}

	int ans = 0;
	for (char a='a'; a<='z'; ++a)	{
		for (char b='a'; b<='z'; ++b)	{
			int tmp = 0;
			for (int i=1; i<=n; ++i)	{
				if (!ok[i])	continue;
				bool flag = true;
                for (int j=0; j<vec[i].size (); ++j)	{
					char c = vec[i][j];
					if (c != a && c != b)	{
						flag = false;
					}
                }
                if (flag)	tmp += len[i];
			}
			ans = max (ans, tmp);
		}
	}

	printf ("%d
", ans);


   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}

　

sorting B - Anton and Lines

题意：给了一些直线，问是否在横坐标(x1, x2)范围内有相交的点

分析：很好想到每条直线与x = x1以及x = x2的直线的交点，那么满足相交的条件是y11 < y12 && y12 > y22，也就是逆序对。这样少掉了正好在x1或x2相交的情况，一种方法是L += EPS, R -= EPS，还有一种是排序。还有升级版的问题

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 21:33:17
* File Name     :B_2.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
struct Y	{
    double y1, y2;
    bool operator < (const Y &r) const	{
		return y1 < r.y1;
    }
}y[N];
/*
    快速读入输出(读入输出外挂)!--黑科技
    使用场合：huge input (1e6以上)
*/
inline int read(void)   {
    int f = 1, ret = 0; char ch = getchar ();
    while ('0' > ch || ch > '9')    {
        if (ch == '-')  f = -1;
        ch = getchar ();
    }
    while ('0' <= ch && ch <= '9')  {
        ret = ret * 10 + ch - '0';
        ch = getchar ();
    }
    return ret * f;
}

int main(void)    {
	int n;	scanf ("%d", &n);
	double x1, x2;	scanf ("%lf%lf", &x1, &x2);
	x1 += EPS;	x2 -= EPS;
	double k, b;
	for (int i=1; i<=n; ++i)	{
		scanf ("%lf%lf", &k, &b);
		y[i].y1 = k * x1 + b;
		y[i].y2 = k * x2 + b;
	}
	sort (y+1, y+1+n);
	bool flag = false;
	for (int i=2; i<=n; ++i)	{
		if (y[i].y2 < y[i-1].y2)		{
			flag = true;	break;
		}
	}
	if (flag)	puts ("YES");
	else	puts ("NO");

//	cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}

　　

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 21:33:17
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
struct Y	{
    ll y1, y2;
    bool operator < (const Y &r) const	{
		return y1 < r.y1 || (y1 == r.y1 && y2 < r.y2);
    }
}y[N];
/*
    快速读入输出(读入输出外挂)!--黑科技
    使用场合：huge input (1e6以上)
*/
inline int read(void)   {
    int f = 1, ret = 0; char ch = getchar ();
    while ('0' > ch || ch > '9')    {
        if (ch == '-')  f = -1;
        ch = getchar ();
    }
    while ('0' <= ch && ch <= '9')  {
        ret = ret * 10 + ch - '0';
        ch = getchar ();
    }
    return ret * f;
}

int main(void)    {
	int n;	scanf ("%d", &n);
	int x1, x2;	scanf ("%d%d", &x1, &x2);
	ll k, b;
	for (int i=1; i<=n; ++i)	{
		k = read ();	b = read ();
		y[i].y1 = k * x1 + b;
		y[i].y2 = k * x2 + b;
	}
	sort (y+1, y+1+n);
	bool flag = false;
	for (int i=2; i<=n; ++i)	{
		if (y[i].y2 < y[i-1].y2)		{
			flag = true;	break;
		}
	}
	if (flag)	puts ("YES");
	else	puts ("NO");

//	cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}