Wannafly挑战赛14

Posted Algorithms Crush Me

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A.角三棱锥

枚举推式子

技术分享图片
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 LL gcd(LL a, LL b){
 5     return a % b ? gcd(b, a % b) : b;
 6 }
 7 int main(){
 8     int T;
 9     scanf("%d", &T);
10     while(T--) {
11         LL K, M, six = 6;
12         cin >> K >> M;
13         LL a = K + 1, b = K + 2, c = K + 3;
14         LL g = gcd(a, six);
15         a /= g, six /= g;
16         g = gcd(b, six);
17         b /= g, six /= g;
18         g = gcd(c, six);
19         c /= g, six /= g;
20         LL ans = a * b % M * c % M;
21         cout << ans << endl;
22     }
23     return 0;
24 }
Aguin

 

B.缀查询

子树修改子树询问转路径标记

技术分享图片
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e6 + 10;
 4 typedef long long LL;
 5  
 6 int node;
 7 LL val[maxn], num[maxn], del[maxn];
 8 LL sum[maxn], tot[maxn];
 9 map<char, int> G[maxn];
10 void insert(string s, LL v){
11     int p = 0;
12     for(int i = 0; i < s.length(); ++i){
13         if(G[p].find(s[i]) == G[p].end()) G[p][s[i]] = ++node;
14         p = G[p][s[i]];
15         v -= del[p];
16         sum[p] += v;
17         tot[p]++;
18     }
19     val[p] += v;
20     num[p]++;
21 }
22 void modify(string s, LL v) {
23     int p = 0;
24     for (int i = 0; i < s.length(); ++i) {
25         if (G[p].find(s[i]) == G[p].end()) G[p][s[i]] = ++node;
26         p = G[p][s[i]];
27     }
28     del[p] += v;
29     LL tmp = tot[p];
30     p = 0;
31     for (int i = 0; i < s.length(); ++i) {
32         if (G[p].find(s[i]) == G[p].end()) G[p][s[i]] = ++node;
33         p = G[p][s[i]];
34         if(i != s.length() - 1) sum[p] += tmp * v;
35     }
36 }
37 LL Q1(string s) {
38     int p = 0;
39     LL tmp = 0;
40     for (int i = 0; i < s.length(); ++i) {
41         if (G[p].find(s[i]) == G[p].end()) G[p][s[i]] = ++node;
42         p = G[p][s[i]];
43         tmp += del[p];
44     }
45     return val[p] + tmp * num[p];
46 }
47 LL Q2(string s) {
48     int p = 0;
49     LL tmp = 0;
50     for (int i = 0; i < s.length(); ++i) {
51         if (G[p].find(s[i]) == G[p].end()) G[p][s[i]] = ++node;
52         p = G[p][s[i]];
53         tmp += del[p];
54     }
55     return sum[p] + tmp * tot[p];
56 }
57  
58 char s[maxn];
59 int main(){
60     int N;
61     scanf("%d", &N);
62     for(int i = 1; i <= N; ++i){
63         int o, a, b;
64         scanf("%d", &o);
65         if(o == 1){
66             scanf("%s %d", s, &a);
67             insert(string(s), a);
68         }
69         else if(o == 2){
70             scanf("%s %d", s, &a);
71             modify(string(s), a);
72         }
73         else if(o == 3){
74             scanf("%s", s);
75             printf("%lld\n", Q1(string(s)));
76         }
77         else{
78             scanf("%s", s);
79             printf("%lld\n", Q2(string(s)));
80         }
81     }
82     return 0;
83 }
Aguin

 

C.达性

缩点

技术分享图片
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 
 5 // BCC
 6 set<int> SE[maxn], ans;
 7 set<int> :: iterator it;
 8 stack<int> S;
 9 vector<int> G[maxn];
10 int dfs_clock, dfn[maxn], low[maxn];
11 int bcc_cnt, bccno[maxn];
12 void dfs(int u, int fa)
13 {
14     dfn[u] = low[u] = ++dfs_clock;
15     S.push(u);
16     for(int i = 0; i < G[u].size(); ++i)
17     {
18         int v = G[u][i];
19         if(!dfn[v])
20         {
21             dfs(v, u);
22             low[u] = min(low[u], low[v]);
23         }
24         else if(!bccno[v]) low[u] = min(low[u], dfn[v]);
25     }
26 
27     if(low[u] == dfn[u])
28     {
29         bcc_cnt++;
30         while(1)
31         {
32             int x = S.top(); S.pop();
33             bccno[x] = bcc_cnt;
34             SE[bcc_cnt].insert(x);
35             if(x == u) break;
36         }
37     }
38 }
39 
40 void find_bcc(int n)
41 {
42     memset(dfn, 0, sizeof(dfn));
43     memset(bccno, 0, sizeof(bccno));
44     dfs_clock = bcc_cnt = 0;
45     for(int i = 1; i <= n; i++) if(!dfn[i]) dfs(i, 0);
46 }
47 
48 int deg[maxn];
49 int main(){
50     int n, m;
51     scanf("%d %d", &n, &m);
52     for(int i = 1; i <= m; ++i){
53         int u, v;
54         scanf("%d %d", &u, &v);
55         G[u].push_back(v);
56     }
57     find_bcc(n);
58     for(int i = 1; i <= n; ++i){
59         for(int j = 0; j < G[i].size(); ++j)
60             if(bccno[i] != bccno[G[i][j]]) deg[bccno[G[i][j]]]++;
61     }
62     for(int i = 1; i <= bcc_cnt; ++i){
63         if(deg[i] == 0) ans.insert(*SE[i].begin());
64     }
65     printf("%d\n", ans.size());
66     for(it = ans.begin(); it != ans.end(); it++){
67         if(it != ans.begin()) putchar( );
68         printf("%d", *it);
69     }
70     puts("");
71     return 0;
72 }
Aguin

 

D.codeJan和树

启发式合并

技术分享图片
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 typedef long long LL;
 5 typedef pair<int, int> pii;
 6 vector<pii> G[maxn];
 7 int n, m;
 8 
 9 int sz[maxn];
10 LL bt[maxn];
11 void dfs(int x, int f){
12     bt[x] = 0;
13     sz[x] = 1;
14     for(int i = 0; i < G[x].size(); ++i){
15         int to = G[x][i].first, d = G[x][i].second;
16         if(to == f) continue;
17         dfs(to, x);
18         bt[x] += bt[to] + d * sz[to];
19         sz[x] += sz[to];
20     }
21 }
22 
23 LL ans;
24 int id[maxn];
25 map<LL, int> S[maxn];
26 map<LL, int> :: iterator it;
27 void add(int x, int f, int y){
28     S[y][bt[x]] = 1;
29     for(int i = 0; i < G[x].size(); ++i){
30         int to = G[x][i].first;
31         if(to == f) continue;
32         add(to, x, y);
33     }
34 }
35 void dfs1(int x, int f){
36     id[x] = x;
37     int M = 0, ms = 0;
38     for(int i = 0; i < G[x].size(); ++i){
39         int to = G[x][i].first;
40         if(to == f) continue;
41         dfs1(to, x);
42         if(sz[to] > M) M = sz[to], ms = to;
43     }
44     if(ms) id[x] = id[ms];
45     for(int i = 0; i < G[x].size(); ++i){
46         int to = G[x][i].first;
47         if(to == f || to == ms) continue;
48         add(to, x, id[x]);
49     }
50     it = S[id[x]].lower_bound(bt[x] - m);
51     if(it != S[id[x]].end()) ans = max(ans, bt[x] - (*it).first);
52     S[id[x]][bt[x]] = 1;
53 }
54 
55 int main(){
56     int T;
57     scanf("%d", &T);
58     while(T--){
59         scanf("%d %d", &n, &m);
60         for(int i = 1; i <= n; ++i) G[i].clear(), S[i].clear();
61         for(int i = 1; i < n; ++i){
62             int u, v, d;
63             scanf("%d %d %d", &u, &v, &d);
64             G[u].push_back(pii(v, d));
65             G[v].push_back(pii(u, d));
66         }
67         ans = -1;
68         dfs(1, 0);
69         dfs1(1, 0);
70         printf("%lld\n", ans);
71     }
72     return 0;
73 }
Aguin

 

E.效位置

倒着并茶几

技术分享图片
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 int a[maxn], b[maxn], ans[maxn];
 5 int base[maxn][33], vis[maxn];
 6 
 7 int fa[maxn];
 8 int Find(int x){
 9     return fa[x] == x ? x : fa[x] = Find(fa[x]);
10 }
11 void Union(int x, int y){
12     x = Find(x), y = Find(y);
13     if(x == y) return;
14     for(int i = 29; i >= 0; i--){
15         if(!base[x][i]) continue;
16         for(int j = 29; j >= 0; j--){
17             if((1 << j) & base[x][i]){
18                 if(base[y][j]) base[x][i] ^= base[y][j];
19                 else {
20                     base[y][j] = base[x][i];
21                     break;
22                 }
23             }
24         }
25     }
26     fa[x] = y;
27 }
28 
29 int main(){
30     int N, M = 0;
31     scanf("%d", &N);
32     for(int i = 1; i <= N; ++i) fa[i] = i;
33     for(int i = 1; i <= N; ++i) scanf("%d", a + i);
34     for(int i = 1; i <= N; ++i) scanf("%d", b + i);
35     for(int i = N; i >= 1; --i){
36         int x = a[b[i]], y = 0;
37         for(int j = 29; j >= 0; --j){
38             if((1 << j) & x){
39                 base[b[i]][j] = x;
40                 break;
41             }
42         }
43         vis[b[i]] = 1;
44         if(vis[b[i] - 1]) Union(b[i] - 1, b[i]);
45         if(vis[b[i] + 1]) Union(b[i] + 1, b[i]);
46         x = Find(b[i]);
47         for(int j = 29; j >= 0; --j){
48             if((1 << j) & y) continue;
49             y ^= base[x][j];
50         }
51         M = max(M, y);
52         ans[i] = M;
53     }
54     for(int i = 1; i <= N; ++i) printf("%d\n", ans[i]);
55     return 0;
56 }
Aguin

 

F.

 

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