#include <bits/stdc++.h> using namespace std; const int N = 60; char name[N][N]; map<string, int> mp; char s[N]; struct P { int id, point; int dif, goal; bool operator < (const P &rhs) const { if (point == rhs.point && goal - dif == rhs.goal - rhs.dif) return goal > rhs.goal; if (point == rhs.point) return goal - dif > rhs.goal - rhs.dif; return point > rhs.point; } } p[N]; int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%s", name[i]); mp[name[i]] = i; p[i].id = i; } for (int i = 0; i < n * (n - 1) / 2; i++) { int pp, qq; scanf("%s%d:%d", s, &pp, &qq); int j = 0; int len = strlen(s); char s1[50] = {}; char s2[50] = {}; int p1 = 0, p2 = 0; for (; s[j] != '-'; j++) s1[p1++] = s[j]; j++; for (; j < len; j++) s2[p2++] = s[j]; int a = mp[s1], b = mp[s2]; p[a].goal += pp, p[b].goal += qq; p[a].dif += qq; p[b].dif += pp; if (pp > qq) p[a].point += 3; else if (pp == qq) p[a].point++, p[b].point++; else p[b].point += 3; } sort(p, p + n); vector<string> ans; for (int i = 0; i < n / 2; i++) ans.push_back(name[p[i].id]); sort(ans.begin(), ans.end()); for (auto x: ans) cout << x << '\n'; return 0; }
有$n$个商品，每个商品价格$c_i$，售货员处理这件商品需要时间$t_i$，偷走一个商品需要$1$个单位时间，问怎么安排这些商品的结账顺序可以使的最后要还的钱最少。
如果让一件商品去结账，相当于花$c_i$块钱，能偷走买走至多$t_i + 1$个商品，要使最后总花费最小，那么就是01背包了。
#include <bits/stdc++.h> #define ll long long using namespace std; const int N = 2020; ll dp[N]; int w[N]; ll c[N]; int main() { memset(dp, 0x3f, sizeof dp); dp[0] = 0; int n; scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d%lld", &w[i], &c[i]), w[i]++; for (int i = 1; i <= n; i++) { for (int j = n; j >= 1; j--) { dp[j] = min(dp[j], dp[max(j - w[i], 0)] + c[i]); } } printf("%lld\n", dp[n]); return 0; }
有一个序列，每种元素至多出现$10$次，如果有一个子串前一半等于后一半，那么把前一半及以前的字符全部删掉，要求从短的以及较左的地方开始删，问最后这个串长什么样。
#include <bits/stdc++.h> using namespace std; const int N = 1e5 + 7; int a[N], v[N]; vector<int> G[N]; int main() { int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &a[i]); v[i] = a[i]; } sort(v + 1, v + 1 + n); int cnt = unique(v + 1, v + 1 + n) - v - 1; for (int i = 1; i <= n; i++) { a[i] = lower_bound(v + 1, v + 1 + cnt, a[i]) - v; G[a[i]].push_back(i); } int now = 0; for (int i = 2; i <= n; i++) { bool flag = 0; for (auto pos: G[a[i]]) { flag = 1; if (pos >= i) continue; if (n - i + 1 < i - pos) continue; if (pos <= now) continue; flag = 0; for (int j = 1; j < i - pos; j++) { if (a[i + j] != a[pos + j]) { flag = 1; break; } } if (!flag) break; } if (!flag) now = i - 1; } printf("%d\n", n - now); for (int i = now + 1; i <= n; i++) printf("%d%c", v[a[i]], " \n"[i == n]); return 0; }
一个二维坐标系，有三个操作。
1.加点
2.删点
3.查询严格在$(x, y)$右上角的点，输出$x$坐标最小的那个点，如果有相同的，则输出$y$最小的，不存在输出$-1$
因为存在修改操作，那么就不能离线做了，用线段树维护$x$坐标下$y$坐标的最大值，然后就变成了查询$\left[x + 1, inf\right]$中最小的$x$其$y$大于要查询的$y$，那么就又是先查左及剪枝的那个写法了。然后找到对应$x$坐标后，加点删点用一个set维护每个$x$坐标出现过$y$的坐标，在这个set里面lower_bound一下就好了，刚开始都插入0以及所有$y$坐标都加个一会好操作很多。
#include <bits/stdc++.h> using namespace std; const int N = 5e5 + 7; struct Seg { #define lp p << 1 #define rp p << 1 | 1 int mx[N << 2]; void pushup(int p) { mx[p] = max(mx[lp], mx[rp]); } void update(int p, int l, int r, int x, int val) { if (l == r) { mx[p] = max(val, mx[p]); return; } int mid = l + r >> 1; if (x <= mid) update(lp, l, mid, x, val); else update(rp, mid + 1, r, x, val); pushup(p); } void change(int p, int l, int r, int x, int val) { if (l == r) { mx[p] = val; return; } int mid = l + r >> 1; if (x <= mid) change(lp, l, mid, x, val); else change(rp, mid + 1, r, x, val); pushup(p); } int query(int p, int l, int r, int x, int y, int val) { if (x > y) return -1; if (mx[p] <= val) return -1; if (l == r) return l; int mid = l + r >> 1; if (x <= mid) { int temp = query(lp, l, mid, x, y, val); if (temp != -1) return temp; } return query(rp, mid + 1, r, x, y, val); } } seg; struct OPT { int opt; int x, y; } o[N]; int v[N]; set<int> st[N]; int main() { // freopen("in.txt", "r", stdin); int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { char s[10] = {}; scanf("%s%d%d", s, &o[i].x, &o[i].y); if (s[0] == 'a') o[i].opt = 0; else if (s[0] == 'r') o[i].opt = 1; else o[i].opt = 2; v[i] = o[i].x; o[i].y++; } sort(v + 1, v + 1 + n); int cnt = unique(v + 1, v + 1 + n) - v - 1; for (int i = 1; i <= cnt; i++) st[i].insert(0); for (int i = 1; i <= n; i++) o[i].x = lower_bound(v + 1, v + 1 + cnt, o[i].x) - v; for (int i = 1; i <= n; i++) { if (o[i].opt == 0) { st[o[i].x].insert(o[i].y); seg.update(1, 1, cnt, o[i].x, o[i].y); } else if (o[i].opt == 1) { st[o[i].x].erase(o[i].y); set<int>::iterator it = st[o[i].x].end(); it--; int res = 0; res = *it; seg.change(1, 1, cnt, o[i].x, res); } else { int pos = seg.query(1, 1, cnt, o[i].x + 1, cnt, o[i].y); if (pos == -1) printf("-1\n"); else { int ans = *st[pos].upper_bound(o[i].y); printf("%d %d\n", v[pos], ans - 1); } } } return 0; }