A. Lunar New Year and Cross Counting
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 510 5 int n; 6 char G[N][N]; 7 8 int main() 9 { 10 while (scanf("%d", &n) != EOF) 11 { 12 for (int i = 1; i <= n; ++i) scanf("%s", G[i] + 1); 13 int res = 0; 14 for (int i = 2; i < n; ++i) for (int j = 2; j < n; ++j) if (G[i][j] == 'X') 15 { 16 if (G[i - 1][j + 1] == 'X' && G[i - 1][j - 1] == 'X' && G[i + 1][j - 1] == 'X' && G[i + 1][j + 1] == 'X') 17 ++res; 18 } 19 printf("%d\n", res); 20 } 21 return 0; 22 }
B. Lunar New Year and Food Ordering
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 100010 6 int n, m; 7 struct qnode 8 { 9 int a, c, id; 10 void scan(int id) { scanf("%d%d", &a, &c); this->id = id; } 11 bool operator < (const qnode &other) const 12 { 13 if (c != other.c) return c < other.c; 14 return id < other.id; 15 } 16 }q[N]; 17 18 19 int main() 20 { 21 while (scanf("%d%d", &n, &m) != EOF) 22 { 23 for (int i = 1; i <= n; ++i) scanf("%d", &q[i].a); 24 for (int i = 1; i <= n; ++i) scanf("%d", &q[i].c); 25 for (int i = 1; i <= n; ++i) q[i].id = i; 26 queue <int> que; 27 sort(q + 1, q + 1 + n); 28 for (int i = 1; i <= n; ++i) que.push(q[i].id); 29 sort(q + 1, q + 1 + n, [](qnode a, qnode b) { return a.id < b.id; } ); 30 for (int i = 1, d, t; i <= m; ++i) 31 { 32 scanf("%d%d", &t, &d); 33 ll res = 0; 34 if (q[t].a >= d) 35 { 36 res += 1ll * d * q[t].c; 37 q[t].a -= d; 38 d = 0; 39 } 40 else 41 { 42 res += 1ll * q[t].a * q[t].c; 43 d -= q[t].a; 44 q[t].a = 0; 45 while (!que.empty() && d) 46 { 47 int top = que.front(); 48 if (q[top].a >= d) 49 { 50 res += 1ll * d * q[top].c; 51 q[top].a -= d; 52 d = 0; 53 break; 54 } 55 else 56 { 57 res += 1ll * q[top].c * q[top].a; 58 d -= q[top].a; 59 q[top].a = 0; 60 que.pop(); 61 } 62 } 63 } 64 if (d > 0) res = 0; 65 printf("%lld\n", res); 66 } 67 } 68 return 0; 69 }
C. Lunar New Year and Number Division
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 300010 6 int n; 7 ll a[N]; 8 9 int main() 10 { 11 while (scanf("%d", &n) != EOF) 12 { 13 for (int i = 1; i <= n; ++i) scanf("%lld", a + i); 14 sort(a + 1, a + 1 + n); 15 ll res = 0; 16 for (int i = 1; i <= n >> 1; ++i) 17 { 18 res += 1ll * (a[i] + a[n - i + 1]) * (a[i] + a[n - i + 1]); 19 } 20 printf("%lld\n", res); 21 } 22 return 0; 23 }
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D. Lunar New Year and a Wander
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 int n, m; 6 vector <int> G[N]; 7 vector <int> res; 8 9 int used[N]; 10 void BFS() 11 { 12 priority_queue <int, vector <int>, greater <int> > pq; pq.push(1); 13 res.clear(); 14 memset(used, 0, sizeof used); 15 used[1] = 1; 16 while (!pq.empty()) 17 { 18 res.push_back(pq.top()); 19 int u = pq.top(); pq.pop(); 20 for (auto v : G[u]) if (!used[v]) 21 { 22 used[v] = 1; 23 pq.push(v); 24 } 25 } 26 } 27 28 int main() 29 { 30 while (scanf("%d%d", &n, &m) != EOF) 31 { 32 for (int i = 1; i <= n; ++i) G[i].clear(); 33 for (int i = 1, u, v; i <= m; ++i) 34 { 35 scanf("%d%d", &u, &v); 36 G[u].push_back(v); 37 G[v].push_back(u); 38 } 39 BFS(); 40 for (int i = 0; i < n; ++i) printf("%d%c", res[i], " \n"[i == n - 1]); 41 } 42 return 0; 43 }
E. Lunar New Year and Red Envelopes
Solved.
题意：
有$n个红包，每个红包可以在[s_i, t_i]的时间内拾起，拾起后获得w_i元，但是到d_i这个时刻都不能再获取其他红包$
$a非常贪心，他在每一时刻，如果这个时刻可以拾起红包，它会拾起金额最大的，如果同时有多个金额最大的$
$他会拾起d_i最大的$
$他的女儿可以有m次干扰他的机会，每次干扰他使得他在某个时刻不能拾起红包$
$如果他的女红采用最优操作，他最少获得多少金钱$
思路：
我们发现在某个时刻如果可以拾起红包，那么这个红包已经确定，可以用线段树预处理
再考虑$dp[i][j] 表示在第i个时刻，用了j次机会的最少金币$
$有两个转移$
$一个是 这个时刻如果没有红包可以拿 那么$
$dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])$
如果有红包可以拿
$dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j])$
$dp[d_i + 1][j] = min(dp[d_i + 1][j], dp[i][j] + w_i)$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define INFLL 0x3f3f3f3f3f3f3f3f 6 #define N 100010 7 #define M 210 8 int n, m, k; 9 ll f[N][M]; 10 11 namespace SEG 12 { 13 struct node 14 { 15 int w, d; 16 void init() 17 { 18 w = d = 0; 19 } 20 void add(int ww, int dd) 21 { 22 if (ww > w) 23 { 24 w = ww; 25 d = dd; 26 } 27 else if (ww == w && dd > d) 28 { 29 d = dd; 30 } 31 } 32 }a[N << 2], res; 33 void init() 34 { 35 memset(a, 0, sizeof a); 36 } 37 void pushdown(int id) 38 { 39 a[id << 1].add(a[id].w, a[id].d); 40 a[id << 1 | 1].add(a[id].w, a[id].d); 41 } 42 void update(int id, int l, int r, int ql, int qr, int w, int d) 43 { 44 if (l >= ql && r <= qr) 45 { 46 a[id].add(w, d); 47 return; 48 } 49 int mid = (l + r) >> 1; 50 pushdown(id); 51 if (ql <= mid) update(id << 1, l, mid, ql, qr, w, d); 52 if (qr > mid) update(id << 1 | 1, mid + 1, r, ql, qr, w, d); 53 } 54 void query(int id, int l, int r, int pos) 55 { 56 if (l == r) 57 { 58 res = a[id]; 59 return; 60 } 61 int mid = (l + r) >> 1; 62 pushdown(id); 63 if (pos <= mid) query(id << 1, l, mid, pos); 64 else query(id << 1 | 1, mid + 1, r, pos); 65 } 66 67 } 68 69 int main() 70 { 71 while (scanf("%d%d%d", &n, &m, &k) != EOF) 72 { 73 SEG::init(); 74 memset(f, 0x3f, sizeof f); 75 int last = 0; 76 for (int i =1, s, t, d, w; i <= k; ++i) 77 { 78 scanf("%d%d%d%d", &s, &t, &d, &w); 79 last = max(last, d); 80 SEG::update(1, 1, n, s, t, w, d); 81 } 82 for (int i = 0; i <= m; ++i) f[1][i] = 0; 83 for (int i = 1; i <= n; ++i) 84 { 85 SEG::res.init(); 86 SEG::query(1, 1, n, i); 87 int w = SEG::res.w, d = SEG::res.d; 88 for (int j = 0; j <= m; ++j) 89 { 90 if (d) 91 { 92 f[d + 1][j] = min(f[d + 1][j], f[i][j] + w); 93 f[i + 1][j + 1] = min(f[i + 1][j + 1], f[i][j]); 94 } 95 else 96 f[i + 1][j] = min(f[i + 1][j], f[i][j]); 97 } 98 } 99 ll res = INFLL; 100 for (int i = 0; i <= m; ++i) res = min(res, f[n + 1][i]); 101 printf("%lld\n", res); 102 } 103 return 0; 104 }