A - Beginning
签到.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 int a[4]; 7 while (scanf("%d", a) != EOF) 8 { 9 for (int i = 1; i < 4; ++i) scanf("%d", a + i); 10 sort(a, a + 4); 11 int res = 0; 12 for (int i = 0; i < 4; ++i) res = res * 10 + a[i]; 13 puts(res == 1479 ? "YES" : "NO"); 14 } 15 return 0; 16 } View Code
B - KEYENCE String
签
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 string s; 5 6 string get(int pos) 7 { 8 string res = ""; 9 for (int i = 0; i <= pos; ++i) res += s[i]; 10 int len = s.size(); 11 for (int i = len - (7 - pos - 1); i < len; ++i) res += s[i]; 12 return res; 13 } 14 15 bool work() 16 { 17 for (int i = 0; i < 7; ++i) 18 { 19 string tmp = get(i); 20 if (tmp == "keyence") return 1; 21 } 22 return 0; 23 } 24 25 int main() 26 { 27 while (cin >> s) puts(work() ? "YES" : "NO"); 28 return 0; 29 } View Code
C - Exam and Wizard
Solved.
题意：
给出些个数字$A_i, 和 B_i$
要求构造$C_i 使得 C_i >= B_i 并且 \sum A_i = \sum C_i$
并且使得变动的数字个数最少
思路：
先弄出不足的部分，然后取差值最大的$A_i > B_i$ 的部分用堆维护，每次取最大的贡献出来补充
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 #define ll long long 6 int n, a[N], b[N]; 7 8 int main() 9 { 10 while (scanf("%d", &n) != EOF) 11 { 12 for (int i = 1; i <= n; ++i) scanf("%d", a + i); 13 for (int i = 1; i <= n; ++i) scanf("%d", b + i); 14 ll need = 0; 15 priority_queue <int, vector <int>, less <int> > pq; 16 int res = 0; 17 for (int i = 1; i <= n; ++i) 18 { 19 if (b[i] > a[i]) need += b[i] - a[i], ++res; 20 else if (b[i] < a[i]) pq.push(a[i] - b[i]); 21 } 22 while (!pq.empty() && need > 0) 23 { 24 int top = pq.top(); pq.pop(); 25 need -= top; ++res; 26 } 27 if (need > 0) puts("-1"); 28 else printf("%d\n", res); 29 } 30 return 0; 31 } View Code
D - Double Landscape
Upsolved.
题意：
在$n * m的矩阵中填数，每行中最大的数为a_i, 每列中最大的数为b_i$
求填数方案
思路：
对于某个数$x$
如果它存在于多个$a_i 中 或者多个 b_i 中 $
那么无解
再考虑
它既存在于$a_i 也存在于 b_i 中
那么它的位置是确定的$
它只存在于某个$a_i或者某个b_i中$
那么它的方案数就是那一列或者那一行的$a_i$比它大的数的个数
它不存在于某个$a_i或者某个b_i中$
那么它的方案数是 $a_i > a_x 的个数 \cdot b_i > b_x的个数$
但是要减去比它大的数的个数
因为它能填的位置，比它大的数都能填，而且是子集关系
所以要让出一位
但是对于第二种情况不用考虑，因为第二种情况的那个数肯定是当前行或者当前列最大的
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 1100 6 const ll MOD = (ll)1e9 + 7; 7 int n, m, a[N], b[N]; 8 9 bool check() 10 { 11 for (int i = 1; i <= n; ++i) for (int j = i + 1; j <= n; ++j) 12 if (a[i] == a[j]) return false; 13 for (int i = 1; i <= m; ++i) for (int j = i + 1; j <= m; ++j) 14 if (b[i] == b[j]) return false; 15 return true; 16 } 17 18 int main() 19 { 20 while (scanf("%d%d", &n, &m) != EOF) 21 { 22 for (int i = 1; i <= n; ++i) scanf("%d", a + i); 23 for (int i = 1; i <= m; ++i) scanf("%d", b + i); 24 sort(a + 1, a + 1 + n); 25 sort(b + 1, b + 1 + m); 26 if (!check()) puts("0"); 27 else 28 { 29 a[n + 1] = n + 1; 30 b[m + 1] = m + 1; 31 int l[2] = {0, 0}; 32 ll res = 1; 33 for (int i = 1; i <= n * m; ++i) 34 { 35 while (l[0] <= n && a[l[0]] < i) ++l[0]; 36 while (l[1] <= m && b[l[1]] < i) ++l[1]; 37 ll d = 0; 38 if (a[l[0]] == i && b[l[1]] == i) 39 d = 1; 40 else if (a[l[0]] == i && b[l[1]] != i) 41 d = m - l[1] + 1; 42 else if (b[l[1]] == i && a[l[0]] != i) 43 d = n - l[0] + 1; 44 else 45 d = (m - l[1] + 1) * (n - l[0] + 1) - (n * m - i); 46 res = (res * d) % MOD; 47 } 48 printf("%lld\n", res); 49 } 50 } 51 return 0; 52 } View Code
E - Connecting Cities
Upsolved.
题意：
$n个城市，两个城市之间的边权是 |i - j| \cdot D + A_i + A_j$
求最小生成树
思路：
考虑将区间分成两部分
分别为$[l, mid] 和 [mid +1, r]$
考虑
左区间为$f[i] = a[i] - d * i$
右区间为$g[i] = a[i] + d * j$
令$f[0] 和 g[0] 为 Min(f[i]), g[0] 为 Min(g[i])$
那么我们将$(i, j_0), (i_0, j), (i_0, j_0) 连边$
我么考虑$(i, j)这种边一定比上面三种边的边权要大$
那么假设我们需要$(i, j)这种边来增加连通性，那么上面那三种边必定能满足$
所以分治建边 一共有$NlogN级别的边数 $
再做MST
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 200010 6 #define INFLL 0x3f3f3f3f3f3f3f3f 7 int n, m; 8 ll d, a[N]; 9 struct Edge 10 { 11 int u, v; ll w; 12 Edge() {} 13 Edge(int u, int v, ll w) : u(u), v(v), w(w) {} 14 bool operator < (const Edge &other) const { return w < other.w; } 15 }edge[N * 30]; 16 17 void add(int l, int r) 18 { 19 if (l == r) return; 20 int mid = (l + r) >> 1; 21 ll Min = INFLL; int pos = -1; 22 for (int i = l; i <= mid; ++i) 23 { 24 ll f = a[i] - d * i; 25 if (f < Min) 26 { 27 Min = f; 28 pos = i; 29 } 30 } 31 for (int i = mid + 1; i <= r; ++i) 32 edge[++m] = Edge(pos, i, a[pos] + a[i] + d * (i - pos)); 33 Min = INFLL; pos = -1; 34 for (int i = mid + 1; i <= r; ++i) 35 { 36 ll f = a[i] + d * i; 37 if (f < Min) 38 { 39 Min = f; 40 pos = i; 41 } 42 } 43 for (int i = l; i <= mid; ++i) 44 edge[++m] = Edge(pos, i, a[pos] + a[i] + d * (pos - i)); 45 add(l, mid); 46 add(mid + 1, r); 47 } 48 49 int pre[N]; 50 int find(int x) { return pre[x] == 0 ? x : pre[x] = find(pre[x]); } 51 ll Kruskal() 52 { 53 memset(pre, 0, sizeof pre); 54 sort(edge + 1, edge + 1 + m); 55 int cnt = 1; 56 ll res = 0; 57 for (int i = 1; i <= m; ++i) 58 { 59 int u = edge[i].u, v = edge[i].v; ll w = edge[i].w; 60 int fu = find(u), fv = find(v); 61 if (fu == fv) continue; 62 pre[fu] = fv; 63 res += w; 64 ++cnt; 65 if (cnt == n) return res; 66 } 67 return res; 68 } 69 70 int main() 71 { 72 while (scanf("%d%lld", &n, &d) != EOF) 73 { 74 m = 0; 75 for (int i = 1; i <= n; ++i) scanf("%lld", a + i); 76 add(1, n); 77 printf("%lld\n", Kruskal()); 78 } 79 return 0; 80 } View Code
法二：
按$a_i大小排序$
每次对于当前$x, 在所有a_j <= a_x的 城市当中，对于在它两侧的，各选出一条最短边连边$
为什么这样贪心是对的
我们考虑假如存在一个城市$z\; a_z < a_i, 并且假设最短边的城市是y$
那么$(x, y) 和 (y, z) 都小于 (x, z)$
所以$(x, z)这条边一定不在MST中$
考虑这样证明
$dist(x, y) < dist(x, z) 非常显然$
考虑$dist(y, z) < dist(x, z)$
首先假设$z < y < x$
$dist(y, z) = a_y +a_z + d * (y - z)$
$dist(x, z) = a_x + a_z + d * (x - z)$
$dist(x, z) - dist(y, z) = a_x - a_y + d * (x - y) >= 0$
得证
再考虑$y < z < x$
$dist(x, y) = a_x + a_y + d * (x - y)$
$dist(z, y) = a_z + a_y + d * (z - y)$
$dist(x, y) - dist(z, y) = a_x - a_z + d * (x - z) >= 0$
又因为 $dist(x, y) <= dist(x, z)$
所以$dist(x, z) >= dist(x, y) >= dist(z, y)$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 200010 6 #define INFLL 0x3f3f3f3f3f3f3f3f 7 int n, m; 8 ll d; 9 struct node 10 { 11 int id; ll v; 12 void scan(int id) 13 { 14 this->id = id; 15 scanf("%lld", &v); 16 } 17 bool operator < (const node &other) const { return v < other.v; } 18 }a[N]; 19 struct Edge 20 { 21 int u, v; ll w; 22 Edge () {} 23 Edge (int u, int v, ll w) : u(u), v(v), w(w) {} 24 bool operator < (const Edge &other) const { return w < other.w; } 25 }edge[N << 2]; 26 27 struct SEG 28 { 29 struct node 30 { 31 ll Min; int pos; 32 node () {} 33 node (ll Min, int pos) : Min(Min), pos(pos) {} 34 void init() { Min = INFLL, pos = -1; } 35 node operator + (const node &other) const 36 { 37 node res; res.init(); 38 if (Min < other.Min) 39 { 40 res.Min = Min; 41 res.pos = pos; 42 } 43 else 44 { 45 res.Min = other.Min; 46 res.pos = other.pos; 47 } 48 return res; 49 } 50 }a[N << 2], res; 51 void build(int id, int l, int r) 52 { 53 a[id].init(); 54 if (l == r) return; 55 int mid = (l + r) >> 1; 56 build(id << 1, l, mid); 57 build(id << 1 | 1, mid + 1, r); 58 } 59 void update(int id, int l, int r, int pos, ll v) 60 { 61 if (l == r) 62 { 63 a[id] = node(v, pos); 64 return; 65 } 66 int mid = (l + r) >> 1; 67 if (pos <= mid) update(id << 1, l, mid, pos, v); 68 else update(id << 1 | 1, mid + 1, r, pos, v); 69 a[id] = a[id << 1] + a[id << 1 | 1]; 70 } 71 void query(int id, int l, int r, int ql, int qr) 72 { 73 if (qr < ql) return; 74 if (l >= ql && r <= qr) 75 { 76 res = res + a[id]; 77 return; 78 } 79 int mid = (l + r) >> 1; 80 if (ql <= mid) query(id << 1, l, mid, ql, qr); 81 if (qr > mid) query(id << 1 | 1, mid + 1, r, ql, qr); 82 } 83 }seg[2]; 84 85 int pre[N]; 86 int find(int x) { return pre[x] == 0 ? x : pre[x] = find(pre[x]); } 87 ll Kruskal() 88 { 89 memset(pre, 0, sizeof pre); 90 sort(edge + 1, edge + 1 + m); 91 int cnt = 1; 92 ll res = 0; 93 for (int i = 1; i <= m; ++i) 94 { 95 int u = edge[i].u, v = edge[i].v; ll w = edge[i].w; 96 int fu = find(u), fv = find(v); 97 if (fu == fv) continue; 98 pre[fu] = fv; 99 ++cnt; 100 res += w; 101 if (cnt == n) return res; 102 } 103 return res; 104 } 105 106 int main() 107 { 108 while (scanf("%d%lld", &n, &d) != EOF) 109 { 110 m = 0; 111 for (int i = 1; i <= n; ++i) a[i].scan(i); 112 for (int i = 0; i < 2; ++i) seg[i].build(1, 1, n); 113 sort(a + 1, a + 1 + n); 114 for (int i = 1; i <= n; ++i) 115 { 116 int u = a[i].id, v; 117 for (int j = 0; j < 2; ++j) 118 seg[j].res.init(); 119 seg[0].query(1, 1, n, 1, u - 1); 120 seg[1].query(1, 1, n, u + 1, n); 121 for (int j = 0; j < 2; ++j) if (seg[j].res.pos != -1) 122 { 123 v = seg[j].res.pos; 124 edge[++m] = Edge(u, v, a[i].v + 1ll * (j ? -1 : 1) * d * u + seg[j].res.Min); 125 } 126 seg[0].update(1, 1, n, u, a[i].v - d * u); 127 seg[1].update(1, 1, n, u, a[i].v + d * u); 128 } 129 printf("%lld\n", Kruskal()); 130 } 131 return 0; 132 } View Code
转载于:https://www.cnblogs.com/Dup4/p/10265066.html
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