A - Age of Moyu
题意：给出一张图，从1走到n，如果相邻两次走的边的权值不同，花费+1， 否则花费相同，求最小花费
思路：用set记录有当前点的最小花费有多少种方案到达，然后最短路
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 const int INF = 0x3f3f3f3f;
6 const int maxn = 2e5 + 10;
7
8 struct Edge{
9 int to, nxt, val;
10 Edge(){}
11 Edge(int to, int nxt, int val):to(to), nxt(nxt), val(val){};
12 }edge[maxn << 1];
13
14 set<int>s[maxn];
15 int head[maxn], tot;
16
17 void Init(int n)
18 {
19 for(int i = 0; i <= n; ++i) head[i] = -1, s[i].clear();
20 tot = 0;
21 }
22
23 void addedge(int u, int v,int val)
24 {
25 edge[tot] = Edge(v, head[u], val);head[u] = tot++;
26 }
27
28 struct qnode{
29 int v, c;
30 int pre;
31 int fa;
32 qnode(){}
33 qnode(int v, int c, int pre, int fa) :v(v), c(c), pre(pre), fa(fa){}
34 bool operator < (const qnode &r) const
35 {
36 return c > r.c;
37 }
38 };
39
40 int n, m;
41 int dist[maxn];
42
43 void BFS(int st)
44 {
45 for(int i = 1; i <= n; ++i) dist[i] = INF;
46 priority_queue<qnode>q;
47 dist[st] = 0;
48 q.push(qnode(st, 0, 0, 0));
49 while(!q.empty())
50 {
51 qnode tmp = q.top();
52 q.pop();
53 int u = tmp.v;
54 if(tmp.c > dist[u]) continue;
55 else if(tmp.c == dist[u])
56 {
57 if(s[u].find(tmp.pre) != s[u].end()) continue;
58 s[u].insert(tmp.pre);
59 }
60 else
61 {
62 dist[u] = tmp.c;
63 s[u].clear();
64 s[u].insert(tmp.pre);
65 }
66 for(int i = head[u]; ~i; i = edge[i].nxt)
67 {
68 int v = edge[i].to;
69 int cost = edge[i].val;
70 if(v == tmp.fa) continue;
71 if(dist[u] + (cost != tmp.pre) <= dist[v])
72 {
73 dist[v] = dist[u] + (cost != tmp.pre);
74 if(v != n)
75 {
76 q.push(qnode(v, dist[v], cost, u));
77 }
78 }
79 }
80 }
81 }
82
83 int main()
84 {
85 int t;
86 while(scanf("%d %d", &n, &m) != EOF)
87 {
88 Init(n);
89 for(int i = 1; i <= m; ++i)
90 {
91 int u, v, w;
92 scanf("%d %d %d", &u, &v ,&w);
93 addedge(u, v, w);
94 addedge(v, u, w);
95 }
96 BFS(1);
97 if(dist[n] == INF) dist[n] = -1;
98 printf("%d\n", dist[n]);
99 }
100 return 0;
101 }
B - AraBellaC
留坑。
C - YJJ’s Stack
留坑。
D - Go to school
留坑。
E - GuGuFishtion
留坑。
F - Lord Li's problem
留坑。
G - Reverse Game
留坑。
H - Traffic Network in Numazu
题意：两种操作，第一种是更改一条边权的值，第二种是查询x-y的最短路径，给出的是一颗树加一条边
思路：将形成环的边单独拿出来考虑，那么考虑是否经过这条边使得答案更优，修改操作用线段树或者树状数组维护
1 #include<bits/stdc++.h>
2
3 using namespace std;
4
5 typedef long long ll;
6
7 const int maxn = 1e5 + 10;
8 const int DEG = 20;
9
10 struct EDGE{
11 int u, v, w;
12 }EDGE[maxn], TEMP;
13
14 struct Edge{
15 int to, nxt, val;
16 Edge(){}
17 Edge(int to, int nxt, int val) :to(to), nxt(nxt), val(val){}
18 }edge[maxn << 1];
19
20 int head[maxn], tot, cnt;
21 int father[maxn];
22
23 void Init(int n)
24 {
25 for(int i = 0; i <= n; ++i) head[i] = -1, father[i] = i;
26 tot = cnt = 0;
27 }
28
29 int find(int x)
30 {
31 return x == father[x] ? father[x] : father[x] = find(father[x]);
32 }
33
34 void mix(int x,int y)
35 {
36 x = find(x), y = find(y);
37 if(x != y) father[x] = y;
38 }
39
40 bool same(int x,int y)
41 {
42 return find(x) == find(y);
43 }
44
45 void addedge(int u, int v, int val)
46 {
47 edge[tot] = Edge(v, head[u], val); head[u] = tot++;
48 }
49
50 int dro[maxn];
51 int ord[maxn], son[maxn];
52 int fa[maxn][DEG];
53 int deg[maxn];
54
55 void DFS(int u, int pre)
56 {
57 ord[u] = ++cnt;
58 dro[cnt] = u;
59 for(int i = 1; i < DEG; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
60 for(int i = head[u]; ~i; i = edge[i].nxt)
61 {
62 int v = edge[i].to;
63 if(v == pre) continue;
64 fa[v][0] = u;
65 deg[v] = deg[u] + 1;
66 DFS(v, u);
67 }
68 son[u] = cnt;
69 }
70
71 int LCA(int u, int v)
72 {
73 if(deg[u] > deg[v]) swap(u, v);
74 int hu = deg[u], hv = deg[v];
75 int tu = u, tv = v;
76 for(int det = hv - hu, i = 0; det; det >>= 1, ++i)
77 {
78 if(det & 1)
79 {
80 tv = fa[tv][i];
81 }
82 }
83 if(tu == tv) return tu;
84 for(int i = DEG - 1; i >= 0; --i)
85 {
86 if(fa[tu][i] == fa[tv][i]) continue;
87 tu = fa[tu][i];
88 tv = fa[tv][i];
89 }
90 return fa[tu][0];
91 }
92
93 struct node{
94 int l, r;
95 ll val, lazy;
96 node(){}
97 node(int l,int r, ll val, ll lazy) :l(l), r(r), val(val), lazy(lazy){}
98 }tree[maxn << 2];
99
100 void pushup(int id)
101 {
102 tree[id].val = tree[id << 1].val + tree[id << 1 | 1].val;
103 }
104
105 void pushdown(int id)
106 {
107 if(tree[id].lazy)
108 {
109 ll lazy = tree[id].lazy;
110 tree[id << 1].val += lazy;
111 tree[id << 1 | 1].val += lazy;
112 tree[id << 1].lazy += lazy;
113 tree[id << 1 | 1].lazy += lazy;
114 tree[id].lazy = 0;
115 }
116 }
117
118 void build(int id, int l, int r)
119 {
120 tree[id] = node(l, r, 0, 0);
121 if(l == r) return ;
122 int mid = (l + r) >> 1;
123 build(id << 1, l, mid);
124 build(id << 1 | 1, mid + 1, r);
125 }
126
127 void update(int id, int l, int r, ll val)
128 {
129 if(l <= tree[id].l && r >= tree[id].r)
130 {
131 tree[id].val += val;
132 tree[id].lazy += val;
133 return ;
134 }
135 pushdown(id);
136 int mid = (tree[id].l + tree[id].r) >> 1;
137 if(mid >= l) update(id << 1, l, r, val);
138 if(r > mid) update(id << 1 | 1, l, r, val);
139 pushup(id);
140 }
141
142 ll query(int id, int pos)
143 {
144 if(tree[id].l == pos && tree[id].r == pos)
145 {
146 return tree[id].val;
147 }
148 pushdown(id);
149 int mid = (tree[id].l + tree[id].r) >> 1;
150 if(pos <= mid) return query(id << 1, pos);
151 if(pos > mid) return query(id << 1 | 1, pos);
152 pushup(id);
153 }
154
155 ll getdis(int u,int v)
156 {
157 int root = LCA(u, v);
158 return query(1, ord[u]) + query(1, ord[v]) - 2 * query(1, ord[root]);
159 }
160
161 int n, q;
162
163 int main()
164 {
165 int t;
166 scanf("%d", &t);
167 while(t--)
168 {
169 scanf("%d %d", &n, &q);
170 Init(n);
171 for(int i = 1; i <= n; ++i)
172 {
173 scanf("%d %d %d", &EDGE[i].u, &EDGE[i].v, &EDGE[i].w);
174 int u = EDGE[i].u, v = EDGE[i].v, w = EDGE[i].w;
175 if(same(u, v))
176 {
177 TEMP = EDGE[i];
178 continue;
179 }
180 mix(u, v);
181 addedge(u, v, w);
182 addedge(v, u, w);
183 }
184 fa[1][0] = 1;
185 deg[1] = 0;
186 DFS(1, -1);
187 build(1, 1, n);
188 for(int i = 1; i <= n; ++i)
189 {
190 int u = EDGE[i].u, v = EDGE[i].v, w = EDGE[i].w;
191 if(u == TEMP.u && v == TEMP.v) continue;
192 if(fa[u][0] == v)
193 {
194 update(1, ord[u], son[u], w);
195 }
196 else if(fa[v][0] == u);
197 {
198 update(1, ord[v], son[v], w);
199 }
200 }
201 // for(int i = 1; i <= n; ++i) cout << ord[i] << endl;
202 // for(int i = 1; i <= n; ++i) cout << i << " " << query(1, ord[i]) << endl;
203 while(q--)
204 {
205 int op, x, y;
206 scanf("%d %d %d", &op, &x ,&y);
207 if(op == 0)
208 {
209 int u = EDGE[x].u, v = EDGE[x].v, w = EDGE[x].w;
210 if(u == TEMP.u && v == TEMP.v)
211 {
212 EDGE[x].w = y;
213 TEMP.w = y;
214 }
215 if(fa[u][0] == v)
216 {
217 update(1, ord[u], son[u], y - w);
218 }
219 else if(fa[v][0] == u)
220 {
221 update(1, ord[v], son[v], y - w);
222 }
223 EDGE[x].w = y;
224 }
225 else if(op == 1)
226 {
227 ll ans = getdis(x, y);
228 ans = min(ans, getdis(x, TEMP.u) + TEMP.w + getdis(y, TEMP.v));
229 ans = min(ans, getdis(y, TEMP.u) + TEMP.w + getdis(x, TEMP.v));
230
231 printf("%lld\n", ans);
232 }
233 }
234 }
235 return 0;
236 }
I - Tree
题意：一棵树种，每个点有一个权值，表示可以向上跳几步，两种操作，一种是修改某点权值，还有一种是询问某个点需要跳几次跳出
思路：DFS序分块，弹飞绵阳升级版，注意更新的时候，维护一个$In[u]$ 表示最远跳到块内是第几块，这样就不用多一个log
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 namespace FastIO
5 {
6 #define BUF_SIZE 10000005
7 bool IOerror = false;
8 inline char NC()
9 {
10 static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
11 if (p1 == pend)
12 {
13 p1 = buf;
14 pend = buf + fread(buf, 1, BUF_SIZE, stdin);
15 if (pend == p1)
16 {
17 IOerror = true;
18 return -1;
19 }
20 }
21 return *p1++;
22 }
23
24 inline bool blank(char ch)
25 {
26 return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
27 }
28
29 template <typename T>
30 inline void read(T &x)
31 {
32 char ch;
33 while (blank(ch = NC()));
34 if (IOerror)
35 {
36 x = -1;
37 return;
38 }
39 bool flag = false;
40 if (ch == '-')
41 {
42 flag = true;
43 ch = NC();
44 }
45 if (!isdigit(ch)) while (!isdigit(ch = NC()));
46 for (x = ch - '0'; isdigit(ch = NC()); x = x * 10 + ch - '0');
47 if (flag) x *= -1;
48 }
49
50 void out(int x)
51 {
52 if (x / 10) out(x / 10);
53 putchar(x % 10 + '0');
54 }
55
56 inline void print(int x)
57 {
58 out(x);
59 puts("");
60 }
61 #undef BUF_SIZE
62 }using namespace FastIO;
63
64 #define N 100010
65 #define DEG 20
66 #define block 400
67
68 int t, n, q;
69 int a[N], b[N], In[N], Out[N], fa[N][20], deep[N], p[N], fp[N], cnt;
70 vector <int> G[N];
71
72 void Init()
73 {
74 for (int i = 1; i <= n; ++i) G[i].clear();
75 cnt = 0; fa[1][0] = 1; deep[1] = 0;
76 }
77
78 void DFS(int u)
79 {
80 p[u] = ++cnt;
81 fp[cnt] = u;
82 for (int i = 1; i < DEG; ++i)
83 fa[u][i] = fa[fa[u][i - 1]][i - 1];
84 for (auto v : G[u]) if (v != fa[u][0])
85 {
86 deep[v] = deep[u] + 1;
87 DFS(v);
88 }
89 }
90
91 int GetK(int x, int k)
92 {
93 bitset <20> b; b = k;
94 for (int i = 19; i >= 0; --i) if (b[i])
95 x = fa[x][i];
96 return x;
97 }
98
99 int query(int x)
100 {
101 int res = 0;
102 x = p[x];
103 while (x)
104 {
105 res += b[x];
106 x = Out[x];
107 }
108 return res;
109 }
110
111 void update(int x)
112 {
113 if (a[x] > deep[x])
114 {
115 b[p[x]] = 1;
116 Out[p[x]] = 0;
117 In[p[x]] = p[x];
118 }
119 else
120 {
121 int root = GetK(x, a[x]);
122 if ((p[root] - 1) / block != (p[x] - 1) / block)
123 {
124 b[p[x]] = 1;
125 Out[p[x]] = p[root];
126 In[p[x]] = p[x];
127 }
128 else
129 {
130 b[p[x]] = b[p[root]] + 1;
131 Out[p[x]] = Out[p[root]];
132 In[p[x]] = p[root];
133 }
134 }
135 x = p[x];
136 for (int i = x + 1; i <= n && (i - 1) / block == (x - 1) / block; ++i)
137 {
138 if (In[i] != i)
139 {
140 b[i] = b[In[i]] + 1;
141 Out[i] = Out[In[i]];
142 }
143 }
144 }
145
146 void Run()
147 {
148 read(t);
149 while (t--)
150 {
151 read(n); Init();
152 for (int i = 2; i <= n; ++i)
153 {
154 read(fa[i][0]);
155 G[fa[i][0]].push_back(i);
156 } DFS(1);
157 for (int i = 1; i <= n; ++i) read(a[i]);
158 for (int i = 1; i <= n; ++i)
159 {
160 int x = fp[i];
161 if (a[x] > deep[x])
162 {
163 b[i] = 1;
164 Out[i] = 0;
165 In[i] = i;
166 continue;
167 }
168 int root = GetK(x, a[x]);
169 if ((p[root] - 1) / block != (i - 1) / block)
170 {
171 b[i] = 1;
172 Out[i] = p[root];
173 In[i] = i;
174 }
175 else
176 {
177 b[i] = b[p[root]] + 1;
178 Out[i] = Out[p[root]];
179 In[i] = p[root];
180 }
181 }
182 read(q);
183 for (int i = 1, op, x, v; i <= q; ++i)
184 {
185 read(op); read(x);
186 if (op == 1) print(query(x));
187 else
188 {
189 read(v);
190 a[x] = v;
191 update(x);
192 }
193 }
194 }
195 }
196
197 int main()
198 {
199 #ifdef LOCAL
200 freopen("Test.in", "r", stdin);
201 #endif
202
203 Run();
204 return 0;
205 }
J - Sequence
题意：求$F_n = C * F_{n - 2} + D * F_{n - 1} + \lfloor{\frac {p}{n}}\rfloor$
思路：考虑 最后一项最多有$\sqrt n 项 按这个值分块矩阵快速幂即可$
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 #define ll long long
5
6 const ll MOD = (ll)1e9 + 7;
7
8 int t, n;
9 ll A, B, C, D, P;
10
11 ll Biner(ll x)
12 {
13 ll l = x, r = n, res = l;
14 x = P / x;
15 while (r - l >= 0)
16 {
17 ll mid = (l + r) >> 1;
18 ll tmp = P / mid;
19 if (tmp >= x)
20 {
21 l = mid + 1;
22 res = mid;
23 }
24 else
25 r = mid - 1;
26 }
27 return res;
28 }
29
30 struct node
31 {
32 ll a[3][3];
33 node ()
34 {
35 memset(a, 0, sizeof a);
36 }
37 node operator * (const node &r) const
38 {
39 node ans = node();
40 for (int i = 0; i < 3; ++i) for (int j = 0; j < 3; ++j) for (int k = 0; k < 3; ++k)
41 ans.a[i][j] = (ans.a[i][j] + a[i][k] * r.a[k][j] % MOD) % MOD;
42 return ans;
43 }
44 }base;
45
46 node qmod(node base, int n)
47 {
48 node res = node();
49 res.a[0][0] = B, res.a[0][1] = A; res.a[0][2] = 1;
50 while (n)
51 {
52 if (n & 1) res = res * base;
53 base = base * base;
54 n >>= 1;
55 }
56 return res;
57 }
58
59 ll work()
60 {
61 if (n == 1) return A;
62 if (n == 2) return B;
63 int l = 3, r;
64 while (l <= n)
65 {
66 r = Biner(l);
67 base.a[2][0] = P / l;
68 node res = qmod(base, r - l + 1);
69 B = res.a[0][0], A = res.a[0][1];
70 l = r + 1;
71 }
72 return B;
73 }
74
75 int main()
76 {
77 scanf("%d", &t);
78 while (t--)
79 {
80 scanf("%lld%lld%lld%lld%lld%d", &A, &B, &C, &D, &P, &n);
81 memset(base.a, 0, sizeof base.a);
82 base.a[0][0] = D, base.a[1][0] = C, base.a[0][1] = 1; base.a[2][2] = 1;
83 printf("%lld\n", work());
84 }
85 return 0;
86 }
K - Swordsman
题意：有五种攻击属性，怪物有五种防御属性，所有攻击属性要大于其对应的防御属性便能将其击杀，击杀后有经验加成，求最多杀死多少怪物
思路：用5个set 依次维护即可
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 namespace FastIO
5 {
6 #define BUF_SIZE 10000005
7 bool IOerror = false;
8 inline char NC()
9 {
10 static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
11 if (p1 == pend)
12 {
13 p1 = buf;
14 pend = buf + fread(buf, 1, BUF_SIZE, stdin);
15 if (pend == p1)
16 {
17 IOerror = true;
18 return -1;
19 }
20 }
21 return *p1++;
22 }
23
24 inline bool blank(char ch)
25 {
26 return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
27 }
28
29 template <typename T>
30 inline void read(T &x)
31 {
32 char ch;
33 while (blank(ch = NC()));
34 if (IOerror)
35 {
36 x = -1;
37 return;
38 }
39 bool flag = false;
40 if (ch == '-')
41 {
42 flag = true;
43 ch = NC();
44 }
45 if (!isdigit(ch)) while (!isdigit(ch = NC()));
46 for (x = ch - '0'; isdigit(ch = NC()); x = x * 10 + ch - '0');
47 if (flag) x *= -1;
48 }
49 #undef BUF_SIZE
50 }using namespace FastIO;
51
52 const int maxn = 1e5 + 10;
53
54 struct node{
55 int id;
56 int v;
57 node(){}
58 node(int id, int v) :id(id), v(v){}
59 bool operator < (const node &r) const{
60 return v > r.v;
61 }
62 };
63
64 priority_queue<node>q[10];
65
66 int n, k;
67 int ans[maxn];
68 int arr[maxn][20];
69
70 int main()
71 {
72 int t;
73 read(t);
74 while(t--)
75 {
76 read(n), read(k);
77 for(int i = 1; i <= k; ++i) while(!q[i].empty()) q[i].pop();
78 for(int i = 1; i <= k; ++i) read(ans[i]);
79 for(int i = 1; i <= n; ++i)
80 {
81 for(int j = 1; j <= (k * 2); ++j)
82 {
83 read(arr[i][j]);
84 }
85 q[1].push(node(i, arr[i][1]));
86 }
87 int cnt = 0;
88 while(1)
89 {
90 bool flag = false;
91 for(int i = 1; i <= k; ++i)
92 {
93 while(!q[i].empty())
94 {
95 node tmp = q[i].top();
96 if(tmp.v <= ans[i])
97 {
98 if(i == k)
99 {
100 cnt++;
101 flag = true;
102 for(int j = 1; j <= k; ++j) ans[j] += arr[tmp.id][j + k];
103 }
104 else
105 {
106 tmp.v = arr[tmp.id][i + 1];
107 q[i + 1].push(tmp);
108 }
109 q[i].pop();
110 }
111 else break;
112 }
113 }
114 if(!flag) break;
115 }
116 printf("%d\n", cnt);
117 for(int i = 1; i <= k; ++i) printf("%d%c", ans[i], " \n"[i == k]);
118 }
119 return 0;
120 }
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