Wannafly挑战赛19

Posted 2020-12-02 aguin

A.队列Q

随便on

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 int q[maxn], f[maxn], b[maxn], vis[maxn];
 5 stack<int> F, B;
 6 vector<int> ans;
 7  
 8 int main() {
 9     int N, Q;
10     scanf("%d", &N);
11     for(int i = 1; i <= N; ++i) scanf("%d", q + i);
12     scanf("%d", &Q);
13     while(Q--) {
14         int x;
15         char s[11];
16         scanf("%s %d", s, &x);
17         if(s[0] == ‘F‘) f[x] = 1, b[x] = 0, F.push(x);
18         else f[x] = 0, b[x] = 1, B.push(x);
19     }
20     while(!F.empty()) {
21         int x = F.top(); F.pop();
22         if(!f[x] || vis[x]) continue;
23         vis[x] = 1, ans.push_back(x);
24     }
25     for(int i = 1; i <= N; ++i) {
26         if(b[q[i]] || vis[q[i]]) continue;
27         vis[q[i]] = 1, ans.push_back(q[i]);
28     }
29     while(!B.empty()) {
30         int x = B.top(); B.pop();
31         if(!b[x] || vis[x]) continue;
32         vis[x] = 1, F.push(x);
33     }
34     while(!F.empty()) {
35         int x = F.top(); F.pop();
36         ans.push_back(x);
37     }
38     for(int i = 0; i < N; ++i) printf("%d%c", ans[i], i == N - 1 ? ‘
‘ : ‘ ‘);
39     return 0;
40 }

Aguin

 

B.矩阵

单调队列

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 LL M[505][505], sum[505][505], zero[505][505];
 5 deque<int> dq;
 6  
 7 int main() {
 8     int R, C, X, Y, Z;
 9     scanf("%d %d %d %d %d", &R, &C, &X, &Y, &Z);
10     for(int i = 1; i <= R; ++i) {
11         for(int j = 1; j <= C; ++j) {
12             scanf("%lld", M[i] + j);
13             sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + M[i][j];
14             zero[i][j] = zero[i-1][j] + zero[i][j-1] - zero[i-1][j-1] + (M[i][j] == 0 ? 1 : 0);
15         }
16     }
17     LL ans = 0;
18     for(int U = 1; U <= R; ++U) {
19         for(int D = U; D <= min(R, U + X - 1); ++D) {
20             dq.clear();
21             dq.push_back(0);
22             for(int i = 1; i <= C; ++i) {
23                 while(!dq.empty() && (i - dq.front() > Y || (zero[D][i] - zero[U-1][i]) - (zero[D][dq.front()] - zero[U-1][dq.front()]) > Z)) dq.pop_front();
24                 LL x = sum[D][i] - sum[U-1][i];
25                 if(!dq.empty()) ans = max(ans, x - sum[D][dq.front()] + sum[U-1][dq.front()]);
26                 while(!dq.empty() && x <= sum[D][dq.back()] - sum[U-1][dq.back()]) dq.pop_back();
27                 dq.push_back(i);
28             }
29         }
30     }
31     printf("%lld
", ans);
32     return 0;
33 }

Aguin

 

C.多彩的树

枚举色集容斥

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 5e4 + 10;
 4 typedef long long LL;
 5 const LL mod = 1e9 + 7;
 6 vector<int> G[maxn];
 7 int A[maxn], vis[maxn];
 8 LL cnt, ans[2222];
 9  
10 void dfs(int x, int msk) {
11     vis[x] = 1, cnt++;
12     for(int i = 0; i < G[x].size(); ++i) {
13         int to = G[x][i];
14         if(vis[to] || !((1 << (A[to] - 1)) & msk)) continue;
15         dfs(to, msk);
16     }
17 }
18  
19 int main() {
20     int N, K;
21     scanf("%d %d", &N, &K);
22     for(int i = 1; i <= N; ++i) scanf("%d", A + i);
23     for(int i = 1; i < N; ++i) {
24         int u, v;
25         scanf("%d %d", &u, &v);
26         G[u].push_back(v);
27         G[v].push_back(u);
28     }
29     for(int i = 0; i < (1 << K); ++i) {
30         for(int j = 1; j <= N; ++j) vis[j] = 0;
31         for(int j = 1; j <= N; ++j) {
32             if(!vis[j] && ((1 << (A[j] - 1)) & i)) {
33                 cnt = 0, dfs(j, i);
34                 ans[i] = (ans[i] + cnt * (cnt - 1) / 2 + cnt) % mod;
35             }
36         }
37     }
38     for(int j = 0; j < K; ++j) {
39         for (int i = 0; i < (1 << K); ++i) {
40             if (i & (1 << j)) ans[i] = (ans[i] + mod - ans[i ^ (1 << j)]) % mod;
41         }
42     }
43     LL ret = 0;
44     for(int i = 0; i < (1 << K); ++i) {
45         LL base = 1;
46         for(int j = 0; j < K; ++j) {
47             if(i & (1 << j)) base = base * 131 % mod;
48         }
49         ret = (ret + ans[i] * base) % mod;
50     }
51     printf("%lld
", ret);
52     return 0;
53 }

Aguin

 

D.回文

马拉车前后缀预处理完枚举回文中心

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 int A[26], B[26], p[maxn<<1];
 5 typedef long long LL;
 6 const LL INF = 1e18;
 7 LL L[maxn], R[maxn], SL[maxn], SR[maxn];
 8 char s[maxn], ss[maxn<<1];
 9  
10 int main() {
11     scanf("%s", s + 1);
12     int l = strlen(s + 1);
13     for (int i = 0; i < 26; ++i) scanf("%d %d", A + i, B + i);
14     LL m = 0, sum = 0;
15     for (int i = 1; i <= l; ++i) {
16         sum += A[s[i] - ‘a‘] - B[s[i] - ‘a‘];
17         L[i] = sum - m;
18         SL[i] = SL[i - 1] + A[s[i] - ‘a‘];
19         m = min(m, sum);
20     }
21     m = 0, sum = 0;
22     for (int i = l; i >= 1; --i) {
23         sum += A[s[i] - ‘a‘] - B[s[i] - ‘a‘];
24         R[i] = sum - m;
25         SR[i] = SR[i + 1] + A[s[i] - ‘a‘];
26         m = min(m, sum);
27     }
28     int len = 2 * l + 1;
29     ss[0] = ‘$‘;
30     for (int i = 1; i <= len; i++) {
31         if (i % 2) ss[i] = ‘#‘;
32         else ss[i] = s[i / 2];
33     }
34     ss[len + 1] = ‘^‘;
35     int mx = 0, id;
36     for (int i = 1; i <= len; i++) {
37         if (mx > i) p[i] = min(p[2 * id - i], mx - i);
38         else p[i] = 1;
39         while (ss[i + p[i]] == ss[i - p[i]]) p[i]++;
40         if (p[i] + i > mx) {
41             mx = p[i] + i;
42             id = i;
43         }
44     }
45     LL ans = INF;
46     for (int i = 1; i <= len; ++i) {
47         if (ss[i] == ‘#‘) {
48             if (p[i] == 1) continue;
49             int lb = i / 2 - p[i] / 2 + 1, rb = i / 2 + p[i] / 2;
50             ans = min(ans, SL[lb-1] + SR[rb+1] - max(L[lb-1], R[rb+1]));
51         } else {
52             int lb = i / 2 - p[i] / 2 + 1, rb = i / 2 + p[i] / 2 - 1;
53             ans = min(ans, SL[lb-1] + SR[rb+1] - max(L[lb-1], R[rb+1]));
54         }
55     }
56     printf("%lld
", ans);
57     return 0;
58 }

Aguin

 

E.集合

预处理bfs转移要o1 想不到建图

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 // SPFA_min_cost_flow
  5 const int INF = 1e9;
  6 const int maxn = 1e6;
  7 int dist[555], vis[555];
  8 int pv[555], pe[555];
  9 int cnt, h[555];
 10  
 11 struct edge {
 12     int to, pre, cap, cost;
 13 } e[maxn<<1];
 14  
 15 void init() { //Don‘t forget
 16     cnt = 0;
 17     memset(h, -1, sizeof(h));
 18 }
 19  
 20 void add(int from, int to, int cap, int cost) {
 21     e[cnt].pre = h[from];
 22     e[cnt].to = to;
 23     e[cnt].cap = cap;
 24     e[cnt].cost = cost;
 25     h[from] = cnt;
 26     cnt++;
 27 }
 28  
 29 void ad(int from, int to, int cap, int cost) {
 30     add(from, to, cap, cost);
 31     add(to, from, 0, -cost);
 32 }
 33  
 34 int min_cost_flow(int s, int t) {
 35     int ret = 0;
 36     while (1) {
 37         memset(vis, 0, sizeof(vis));
 38         for (int i = 0; i < 555; i++) dist[i] = INF;
 39         dist[s] = 0;
 40         queue<int> q;
 41         q.push(s);
 42         while (!q.empty()) {
 43             int v = q.front();
 44             q.pop();
 45             vis[v] = 0;
 46             for (int i = h[v]; i >= 0; i = e[i].pre) {
 47                 int to = e[i].to, cap = e[i].cap, cost = e[i].cost;
 48                 if (cap > 0 && dist[to] > dist[v] + cost) {
 49                     pv[to] = v, pe[to] = i;
 50                     dist[to] = dist[v] + cost;
 51                     if (!vis[to]) q.push(to);
 52                     vis[to] = 1;
 53                 }
 54             }
 55         }
 56         if (dist[t] >= 0) return ret; // modify here
 57         int d = INF;
 58         for (int v = t; v != s; v = pv[v])
 59             d = min(d, e[pe[v]].cap);
 60         ret += d * dist[t];
 61         for (int v = t; v != s; v = pv[v]) {
 62             e[pe[v]].cap -= d;
 63             e[pe[v] ^ 1].cap += d;
 64         }
 65     }
 66 }
 67  
 68 char SA[55][22], SB[55][22];
 69 int da[55], db[55], ma[55], mb[55];
 70 vector<int> G[55], D[55];
 71  
 72 queue<int> q;
 73 int dis[1<<16], base[22];
 74  
 75 int get(char * s, int la) {
 76     int ret = 0;
 77     for (int i = 0; i < la; ++i) ret += (s[i] - ‘0‘) * base[i];
 78     return ret;
 79 }
 80  
 81 vector<int> rev[17];
 82 int main() {
 83     for (int i = 0; i < 22; ++i) base[i] = 1 << i;
 84     for (int i = 2; i <= 16; ++i) {
 85         for (int j = 0; j < (1 << i); ++j) {
 86             int o = 0;
 87             for (int k = 0; k < i; ++k)
 88                 if (j & (1 << k)) o += (1 << (i - k - 1));
 89             rev[i].push_back(o);
 90         }
 91     }
 92     int N, M, ans = 0;
 93     scanf("%d", &N);
 94     for (int i = 1; i <= N; ++i) scanf("%s", SA[i]);
 95     scanf("%d", &M);
 96     for (int i = 1; i <= M; ++i) scanf("%s", SB[i]);
 97     for (int i = 1; i <= N; ++i) scanf("%d", da + i), ans += da[i];
 98     for (int i = 1; i <= M; ++i) scanf("%d", db + i), ans += db[i];
 99     for (int i = 1; i <= N; ++i) scanf("%d", ma + i);
100     for (int i = 1; i <= M; ++i) scanf("%d", mb + i);
101     for (int i = 1; i <= N; ++i) {
102         int la = strlen(SA[i]);
103         for (int j = 0; j < (1 << la); ++j) dis[j] = INF;
104         int st = get(SA[i], la);
105         dis[st] = 0;
106         q.push(st);
107         while (!q.empty()) {
108             int x = q.front();
109             q.pop();
110             for (int L = 0; L < la; ++L) {
111                 for (int R = L + 1; R < la; ++R) {
112                     int low = x & (base[L] - 1);
113                     int high = (x >> (R + 1)) << (R + 1);
114                     int mid = rev[R - L + 1][((x - high) >> L)] << L;
115                     int t = low + mid + high;
116                     if (dis[t] == INF) dis[t] = dis[x] + 1, q.push(t);
117                 }
118             }
119         }
120         for (int j = 1; j <= M; ++j)
121             if (strlen(SB[j]) == la && dis[get(SB[j], la)] != INF)
122                 G[i].push_back(j), D[i].push_back(dis[get(SB[j], la)] * min(ma[i], mb[j]) - da[i] - db[j]);
123     }
124     init();
125     int S = N + M + 1, T = S + 1;
126     for (int i = 1; i <= N; ++i) ad(S, i, 1, 0);
127     for (int i = 1; i <= M; ++i) ad(N + i, T, 1, 0);
128     for (int i = 1; i <= N; ++i)
129         for (int j = 0; j < G[i].size(); ++j)
130             ad(i, N + G[i][j], 1, D[i][j]);
131     printf("%d
", ans + min_cost_flow(S, T));
132     return 0;
133 }

Aguin

 

F.K串

哈希莫队

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 3e4 + 10;
 4 char s[maxn];
 5  
 6 int sz, ql[maxn], qr[maxn], id[maxn];
 7 bool cmp(int i, int j) {
 8     if(ql[i] / sz != ql[j] / sz) return ql[i] / sz < ql[j] / sz;
 9     return qr[i] < qr[j];
10 }
11  
12 int sum[26];
13 typedef long long LL;
14 const LL base = 131, mod = 1e9 + 7;
15 vector<LL> v;
16 LL c[maxn], ans[maxn];
17  
18 int cnt[maxn];
19 int main() {
20     int K, Q;
21     scanf("%d %s %d", &K, s + 1, &Q);
22     for(int i = 1; i <= Q; ++i) scanf("%d %d", ql + i, qr + i), id[i] = i;
23     sz = sqrt(Q + 0.5), sort(id + 1, id + 1 + Q, cmp);
24     int l = strlen(s + 1);
25     for(int j = 0; j < 26; ++j) c[0] = (c[0] * base + sum[j]) % mod;
26     v.push_back(c[0]);
27     for(int i = 1; i <= l; ++i) {
28         sum[s[i] - ‘a‘] = (sum[s[i] - ‘a‘] + 1) % K;
29         for(int j = 0; j < 26; ++j) c[i] = (c[i] * base + sum[j]) % mod;
30         v.push_back(c[i]);
31     }
32     sort(v.begin(), v.end());
33     v.erase(unique(v.begin(), v.end()), v.end());
34     for(int i = 0; i <= l; ++i) c[i] = lower_bound(v.begin(), v.end(), c[i]) - v.begin() + 1;
35     LL tmp = 0;
36     int L = 1, R = 0;
37     cnt[c[0]]++;
38     for(int i = 1; i <= Q; ++i) {
39         int x = id[i];
40         while (L < ql[x]) {
41             cnt[c[L-1]]--;
42             tmp -= cnt[c[L-1]];
43             ++L;
44         }
45         while (L > ql[x]) {
46             --L;
47             tmp += cnt[c[L-1]];
48             cnt[c[L-1]]++;
49         }
50         while (R < qr[x]) {
51             ++R;
52             tmp += cnt[c[R]];
53             cnt[c[R]]++;
54         }
55         while (R > qr[x]) {
56             cnt[c[R]]--;
57             tmp -= cnt[c[R]];
58             --R;
59         }
60         ans[x] = tmp;
61     }
62     for(int i = 1; i <= Q; ++i) printf("%lld
", ans[i]);
63     return 0;
64 }

Aguin