Codeforces Round #331 (Div. 2)

 

水 A - Wilbur and Swimming Pool

自从打完北京区域赛,对矩形有种莫名的恐惧..

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#include <bits/stdc++.h> using namespace std;   typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f;   int main(void)  {     int n, x[4], y[4];     cin >> n;     for (int i=0; i<n; ++i)  {         cin >> x[i] >> y[i];     }     if (n == 1) puts ("-1");     else if (n == 2)    {         if (x[0] == x[1] || y[0] == y[1])   puts ("-1");         else    {             int ans = abs (x[0] - x[1]) * abs (y[0] - y[1]);             cout << ans << endl;         }     }     else if (n == 3)    {         int ans = -1;         if (x[0] == x[1])   {             ans = abs (y[0] - y[1]) * abs (x[2] - x[0]);         }         else if (x[0] == x[2])  {             ans = abs (y[0] - y[2]) * abs (x[1] - x[0]);         }         else if (x[1] == x[2])  {             ans = abs (y[1] - y[2]) * abs (x[0] - x[1]);         }         cout << ans << endl;     }     else    {         int ans = -1;         if (x[0] == x[1])   {             ans = abs (y[0] - y[1]) * abs (x[2] - x[0]);         }         else if (x[0] == x[2])  {             ans = abs (y[0] - y[2]) * abs (x[1] - x[0]);         }         else if (x[0] == x[3])  {             ans = abs (y[0] - y[3]) * abs (x[1] - x[0]);         }         cout << ans << endl;     }       return 0; }

 

贪心(水) B - Wilbur and Array

直接扫一边,记录后面的值.忘开long long

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#include <bits/stdc++.h> using namespace std;   typedef long long ll; const int N = 2e5 + 10; const int INF = 0x3f3f3f3f; ll a[N];   int main(void)  {     int n;  cin >> n;     for (int i=1; i<=n; ++i) {         cin >> a[i];     }     ll ans = 0;     ll now = 0, oth = 0;     for (int i=1; i<=n; ++i) {         now = oth;         if (now == a[i])    continue;         else if (now < a[i]) {             ll tmp = a[i] - now;             oth += tmp;             ans += tmp;         }         else    {             ll tmp = now - a[i];             oth -= tmp;             ans += tmp;         }     }     cout << ans << endl;       return 0; }

 

贪心+排序 C - Wilbur and Points

题意:给出n个点以及n个w[i],问一个点的序列,使得w[i] = p[j].y - p[j].x,并且保证不出现p[i].x <= p[j].x && p[i].y <= p[j].y (i > j)

分析:其实就是到水题,map映射下,用set + pair来存储点,自动从小到大排序,最后再判序列是否满足题意.tourist的判断代码好神奇,一直怀疑自己读错题,应该是能查看数据的吧.

+ View Code?

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#include <bits/stdc++.h> using namespace std;   typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; map<int, set<pair<int, int> > >mp; int w[N]; pair<int, int> ans[N];   int main(void)  {     int n;  scanf ("%d", &n);     for (int x, y, i=1; i<=n; ++i)   {         scanf ("%d%d", &x, &y);         mp[y-x].insert (make_pair (x, y));     }     for (int i=1; i<=n; ++i) {         scanf ("%d", &w[i]);     }     for (int i=1; i<=n; ++i) {         if (mp[w[i]].empty ())  {             puts ("NO");    return 0;         }         ans[i] = *(mp[w[i]].begin ());         mp[w[i]].erase (mp[w[i]].begin ());     }     for (int i=2; i<=n; ++i) {         int x1 = ans[i].first, x2 = ans[i-1].first;         int y1 = ans[i].second, y2 = ans[i-1].second;         if (x1 <= x2 && y1 <= y2) {             puts ("NO");    return 0;         }     }     puts ("YES");     for (int i=1; i<=n; ++i) {         int x = ans[i].first;         int y = ans[i].second;         printf ("%d %d\n", x, y);     }       return 0; }