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..
#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
#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.
#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 ", x, y); } return 0; }