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  • 2018 Multi-University Training Contest 3

    好像克拉丽丝小姐姐题解写的超详细我都没啥好说的了

    Problem A. Ascending Rating

    仔细一看m是固定的单调DQ就好了

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 1e7 + 10;
 5 int st, ed, deq[maxn];
 6 int a[maxn];
 7 
 8 inline void add(int i) {
 9     while(st < ed && a[deq[ed]] <= a[i]) ed--;
10     deq[++ed] = i;
11 }
12 
13 int main() {
14     int T;
15     scanf("%d", &T);
16     while(T--) {
17         int n, m, k, p, q, r, MOD;
18         scanf("%d %d %d %d %d %d %d", &n, &m, &k, &p, &q, &r, &MOD);
19         for(int i = 1; i <= k; ++i) scanf("%d", a + i);
20         for(int i = k + 1; i <= n; ++i) a[i] = ((LL) p * a[i - 1] + (LL) q * i + r) % MOD;
21         LL A = 0, B = 0;
22         st = ed = 0;
23         for(int i = n; i >= n - m + 2; --i) add(i);
24         for(int i = n - m + 1; i >= 1; --i) {
25             add(i);
26             while(st < ed && deq[st + 1] > i + m - 1) st++;
27             A += a[deq[st + 1]] ^ i;
28             B += (ed - st) ^ i;
29         }
30         printf("%lld %lld
", A, B);
31     }
32     return 0;
33 }
    Aguin

    Problem B. Cut The String

    回文树好像不太会会阿?

    Problem C. Dynamic Graph Matching

    被卡怕了吓得用$2^{n-2}$枚举三段拼起来……

    还肥肠纸张的用int加了两次没取模！

    还有很多人问这个枚举顺序其实没有影响阿……

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int mod = 1e9 + 7;
 4 int cnt[1<<11], ans[6];
 5 int ppc[1<<11];
 6 
 7 int main() {
 8     for(int i = 0; i < (1 << 11); ++i) ppc[i] = __builtin_popcountll(i) / 2;
 9     int T;
10     scanf("%d", &T);
11     while(T--) {
12         int n, m;
13         scanf("%d %d", &n, &m);
14         for(int i = 0; i < (1 << n); ++i) cnt[i] = 0;
15         cnt[0] = 1;
16         for(int i = 0; i <= n / 2; ++i) ans[i] = 0;
17         for(int i = 1; i <= m; ++i) {
18             char s[11];
19             int u, v;
20             scanf("%s %d %d", s, &u, &v);
21             if(u > v) swap(u, v);
22             int msk1 = (1 << (u - 1)) | (1 << (v - 1));
23             for(int j = 0; j < (1 << (n - v)); ++j) {
24                 for(int k = 0; k < (1 << (v - u - 1)); ++k) {
25                     for(int p = 0; p < (1 << (u - 1)); ++p) {
26                         int msk2 = (j << v) | (k << u) | p;
27                         ans[ppc[msk2|msk1]] = (ans[ppc[msk2|msk1]] + mod - cnt[msk2|msk1]) % mod;
28                         if(s[0] == '+') cnt[msk2|msk1] = (cnt[msk2|msk1] + cnt[msk2]) % mod;
29                         else cnt[msk2|msk1] = (cnt[msk2|msk1] + mod - cnt[msk2]) % mod;
30                         ans[ppc[msk2|msk1]] = (ans[ppc[msk2|msk1]] + cnt[msk2|msk1]) % mod;
31                     }
32                 }
33             }
34             for(int i = 1; i <= n / 2; ++i) printf("%d%c", ans[i], i == n / 2 ? '
' : ' ');
35         }
36     }
37     return 0;
38 }
    Aguin

    Problem D. Euler Function

    谈学姐写的

     1 #include<bits/stdc++.h>
 2 using namespace std;
 3 long long T,i,j,k,n;
 4 int main()
 5 {
 6         scanf("%d",&T);
 7         while (T--)
 8         {
 9                 scanf("%lld",&n);
10                 if (n==1) puts("5");
11                 else printf("%lld
",n+5);
12         }
13 }
    谈学姐

    Problem E. Find The Submatrix

    抄抄题解

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL inf = 1e18;
 5 LL w[101][3001], a[101][3001], f[101][4][2][10001];
 6 int m;
 7 
 8 LL tmp[10001];
 9 void solve(int i, int j, int o, int l, int r, int L, int R) {
10     int mid = (l + r) >> 1, M;
11     tmp[mid] = -inf;
12     for(int x = max(L, mid - m); x <= min(mid, R); ++x)
13         if(tmp[mid] < f[i - 1][j - o][o][x] + a[i][mid - x])
14             tmp[mid] = f[i - 1][j - o][o][x] + a[i][mid - x], M = x;
15     if(l < mid) solve(i, j, o, l, mid - 1, L, M);
16     if(mid < r) solve(i, j, o, mid + 1, r, M, R);
17 }
18 
19 int main() {
20     int T;
21     scanf("%d", &T);
22     while(T--) {
23         int n, A, B;
24         scanf("%d %d %d %d", &n, &m, &A, &B);
25         for(int i = 1; i <= n; ++i)
26             for(int j = 1; j <= m; ++j)
27                 scanf("%lld", &w[i][j]);
28         for(int i = 1; i <= n; ++i) {
29             sort(w[i] + 1, w[i] + 1 + m);
30             a[i][m] = 0;
31             for(int j = m - 1; j >= 0; --j) a[i][j] = a[i][j + 1] + w[i][j + 1];
32         }
33         for(int i = 0; i <= n; ++i)
34             for(int j = 0; j <= B; ++j)
35                 for(int k = 0; k <= 1; ++k)
36                     for(int p = 0; p <= A; ++p)
37                         f[i][j][k][p] = -inf;
38         f[0][0][1][0] = 0;
39         for(int i = 1; i <= n; ++i) {
40             for(int j = 0; j <= B; ++j) {
41                 // f[i][j][0][k] = max{f[i - 1][j][0][x] + a[i][k - x], f[i - 1][j - 1][1][x] + a[i][k - x]}
42                 solve(i, j, 0, 0, A, 0, A);
43                 for(int k = 0; k <= A; ++k) f[i][j][0][k] = max(f[i][j][0][k], tmp[k]);
44                 if(j >= 1) {
45                     solve(i, j, 1, 0, A, 0, A);
46                     for(int k = 0; k <= A; ++k) f[i][j][0][k] = max(f[i][j][0][k], tmp[k]);
47                 }
48                 // f[i][j][1][k] = max{f[i - 1][j][0][k], f[i - 1][j][1][k]}
49                 for(int k = 0; k <= A; ++k) f[i][j][1][k] = max(f[i - 1][j][0][k], f[i - 1][j][1][k]);
50             }
51         }
52         LL ans = 0;
53         for(int i = 0; i <= B; ++i)
54             for(int j = 0; j <= 1; ++j)
55                 for(int k = 0; k <= A; ++k)
56                     ans = max(ans, f[n][i][j][k]);
57         printf("%lld
", ans);
58     }
59     return 0;
60 }
    Aguin

    Problem F. Grab The Tree

    谈学姐教我贪心！

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main() {
 5     int T;
 6     scanf("%d", &T);
 7     while(T--) {
 8         int n, x, sum = 0;
 9         scanf("%d", &n);
10         for(int i = 1; i <= n; ++i) scanf("%d", &x), sum ^= x;
11         for(int i = 1; i < n; ++i) {
12             int u, v;
13             scanf("%d %d", &u, &v);
14         }
15         puts(sum ? "Q" : "D");
16     }
17     return 0;
18 }
    Aguin

    Problem G. Interstellar Travel

    一眼凸包写的时候忘了严格增……

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 200000 + 10;
 4 typedef long long LL;
 5 LL x[maxn], y[maxn], id[maxn], st[maxn];
 6 bool cmp(LL i, LL j) {
 7     if(x[i] != x[j]) return x[i] < x[j];
 8     if(y[i] != y[j]) return y[i] > y[j];
 9     return i < j;
10 }
11 LL cross(LL i, LL j, LL k) {return (x[i] - x[k]) * (y[j] - y[k]) - (x[j] - x[k]) * (y[i] - y[k]);}
12 vector<LL> L;
13 int main() {
14     int T;
15     scanf("%d", &T);
16     while (T--) {
17         int n;
18         scanf("%d", &n);
19         for (int i = 1; i <= n; ++i) scanf("%lld %lld", x + i, y + i), id[i] = i;
20         sort(id + 1, id + 1 + n, cmp);
21         L.clear();
22         for(int i = 1; i <= n; ++i) if(i == 1 || x[id[i]] > x[id[i-1]]) L.push_back(id[i]);
23         int p = 0;
24         for(int i = 0; i < L.size(); ++i) {
25             while (p >= 2 && cross(st[p], L[i], st[p - 1]) > 0) p--;
26             while (p >= 2 && cross(st[p], L[i], st[p - 1]) == 0 && L[i] < st[p]) p--;
27             st[++p] = L[i];
28         }
29         for (int i = 1; i <= p; i++) printf("%lld%c", st[i], i == p ? '
' : ' ');
30     }
31     return 0;
32 }
    Aguin

    Problem H. Monster Hunter

    cmp一直写不对被set去重掉了……

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 typedef long long LL;
 5 LL a[maxn], b[maxn];
 6 vector<int> G[maxn];
 7 
 8 int fa[maxn], vis[maxn];
 9 void dfs(int x, int f) {
10     fa[x] = f;
11     for(int i = 0; i < G[x].size(); ++i) {
12         int to = G[x][i];
13         if(to == f) continue;
14         dfs(to, x);
15     }
16 }
17 
18 int pa[maxn];
19 int Find(int x) {
20     return x == pa[x] ? x : pa[x] = Find(pa[x]);
21 }
22 
23 struct node {
24     int id;
25     LL a, b;
26     node(LL A = 0, LL B = 0, int ID = 0): a(A), b(B), id(ID) {}
27     friend bool operator < (node A, node B) {
28         if(A.b - A.a > 0 && B.b - B.a <= 0) return true;
29         if(A.b - A.a == 0 && B.b - B.a < 0) return true;
30         if(B.b - B.a > 0 && A.b - A.a <= 0) return false;
31         if(B.b - B.a == 0 && A.b - A.a < 0) return false;
32         if(A.b - A.a > 0) {
33             if(A.a != B.a) return A.a < B.a;
34             return A.id < B.id;
35         }
36         if(A.b != B.b) return A.b > B.b;
37         return A.id < B.id;
38     }
39 };
40 set<node> S;
41 
42 int main() {
43     int T;
44     scanf("%d", &T);
45     while(T--) {
46         int n;
47         scanf("%d", &n);
48         for(int i = 1; i <= n; ++i) G[i].clear(), pa[i] = i, vis[i] = 0;
49         S.clear();
50         for(int i = 2; i <= n; ++i)
51             scanf("%lld %lld", a + i, b + i), S.insert(node(a[i], b[i], i));
52         for(int i = 2; i <= n; ++i) {
53             int u, v;
54             scanf("%d %d", &u, &v);
55             G[u].push_back(v), G[v].push_back(u);
56         }
57         dfs(1, 0);
58         LL A = 0, B = 0;
59         for(int i = 1; i < n; ++i) {
60             int x = (*S.begin()).id;
61             S.erase(S.begin());
62             if(fa[x] == 1 || vis[Find(fa[x])]) {
63                 LL t = b[x] + B - a[x] - A;
64                 A = max(A, A - B + a[x]), B = A + t;
65                 vis[x] = 1;
66             }
67             else {
68                 int y = Find(fa[x]);
69                 S.erase(S.find(node(a[y], b[y], y)));
70                 LL t = b[x] + b[y] - a[x] - a[y];
71                 a[y] = max(a[y], a[y] - b[y] + a[x]), b[y] = a[y] + t;
72                 S.insert(node(a[y], b[y], y));
73                 pa[x] = y;
74             }
75         }
76         printf("%lld
", A);
77     }
78     return 0;
79 }
    Aguin

    Problem I. Random Sequence

    感觉还是状态不太好想到吧……

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL mod = 1e9 + 7;
 5 int gcd[101][101], a[101];
 6 LL f[2][101][101][101];
 7 vector<int> fac[101];
 8 LL v[101];
 9 
10 LL fp(LL a, LL b) {
11     LL ret = 1;
12     while (b) {
13         if (b & 1) ret = ret * a % mod;
14         a = a * a % mod;
15         b >>= 1;
16     }
17     return ret;
18 }
19 
20 LL inv(LL x) {
21     return fp(x, mod - 2);
22 }
23 
24 int main() {
25     for(int i = 1; i <= 100; ++i)
26         for(int j = i; j <= 100; j += i)
27             fac[j].push_back(i);
28     for(int i = 1; i <= 100; ++i)
29         for(int j = 1; j <= 100; ++j)
30             gcd[i][j] = __gcd(i, j);
31     int T;
32     scanf("%d", &T);
33     while(T--) {
34         int n, m, c = 0;
35         scanf("%d %d", &n, &m);
36         for(int i = 1; i <= n; ++i) scanf("%d", a + i), c += a[i] == 0;
37         for(int i = 1; i <= m; ++i) scanf("%lld", v + i);
38         int o = 0;
39         for(int i = 1; i <= m; ++i) {
40             for(int j = 0; j < fac[i].size(); ++j) {
41                 for(int k = 0; k < fac[fac[i][j]].size(); ++k) {
42                     f[o][i][fac[i][j]][fac[fac[i][j]][k]] = 0;
43                 }
44             }
45         }
46         for(int i = 1; i <= m; ++i) {
47             if(a[3] && a[3] != i) continue;
48             for(int j = 1; j <= m; ++j) {
49                 if(a[2] && a[2] != j) continue;
50                 for(int k = 1; k <= m; ++k) {
51                     if(a[1] && a[1] != k) continue;
52                     f[o][i][gcd[i][j]][gcd[gcd[i][j]][k]]++;
53                 }
54             }
55         }
56         for(int p = 3; p < n; ++p) {
57             for(int i = 1; i <= m; ++i) {
58                 for(int j = 0; j < fac[i].size(); ++j) {
59                     for(int k = 0; k < fac[fac[i][j]].size(); ++k) {
60                         f[o ^ 1][i][fac[i][j]][fac[fac[i][j]][k]] = 0;
61                     }
62                 }
63             }
64             for(int i = 1; i <= m; ++i) {
65                 if(a[p] && a[p] != i) continue;
66                 for(int j = 0; j < fac[i].size(); ++j) {
67                     if(a[p - 1] && gcd[a[p - 1]][i] != fac[i][j]) continue;
68                     for(int k = 0; k < fac[fac[i][j]].size(); ++k) {
69                         if(a[p - 2] && gcd[fac[i][j]][a[p - 2]] != fac[fac[i][j]][k]) continue;
70                         for(int q = 1; q <= m; ++q) {
71                             if(a[p + 1] && a[p + 1] != q) continue;
72                             f[o ^ 1][q][gcd[q][i]][gcd[q][fac[i][j]]] = (f[o ^ 1][q][gcd[q][i]][gcd[q][fac[i][j]]] + v[gcd[q][fac[fac[i][j]][k]]] * f[o][i][fac[i][j]][fac[fac[i][j]][k]]) % mod;
73                         }
74                     }
75                 }
76             }
77             o ^= 1;
78         }
79         LL ans = 0;
80         for(int i = 1; i <= m; ++i) {
81             for(int j = 0; j < fac[i].size(); ++j) {
82                 for(int k = 0; k < fac[fac[i][j]].size(); ++k) {
83                     ans = (ans + f[o][i][fac[i][j]][fac[fac[i][j]][k]]) % mod;
84                 }
85             }
86         }
87         printf("%lld
", ans * fp(inv(m), c) % mod);
88     }
89     return 0;
90 }
    Aguin

    Problem J. Rectangle Radar Scanner

    快乐分治，好像每次加完点要删掉

      1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long LL;
  4 const int maxn = 1e5 + 10;
  5 const int maxm = 1e6 + 10;
  6 LL xl[maxm], xr[maxm], yl[maxm], yr[maxm], P[maxm], MAX[maxm], MIN[maxm];
  7 LL k, y[maxn], w[maxn];
  8 int n;
  9 
 10 // seg
 11 const LL INF = 1e18;
 12 LL pr[maxn<<2], ma[maxn<<2], mi[maxn<<2];
 13 void gather(int p) {
 14     pr[p] = pr[p << 1] * pr[p << 1 | 1] % k;
 15     ma[p] = max(ma[p << 1], ma[p << 1 | 1]);
 16     mi[p] = min(mi[p << 1], mi[p << 1 | 1]);
 17 }
 18 void build(int p, int l, int r) {
 19     if (l < r) {
 20         int mid = (l + r) >> 1;
 21         build(p << 1, l, mid);
 22         build(p << 1 | 1, mid + 1, r);
 23         gather(p);
 24     } else pr[p] = 1, mi[p] = INF, ma[p] = -INF;
 25 }
 26 void modify(int p, int tl, int tr, int x, LL y) {
 27     if (tl == tr) {
 28         if (!y) pr[p] = 1, mi[p] = INF, ma[p] = -INF;
 29         else pr[p] = pr[p] * y % k, mi[p] = min(mi[p], y), ma[p] = max(ma[p], y);
 30         return;
 31     }
 32     int mid = (tl + tr) >> 1;
 33     if (x <= mid) modify(p << 1, tl, mid, x, y);
 34     else modify(p << 1 | 1, mid + 1, tr, x, y);
 35     gather(p);
 36 }
 37 typedef pair<LL, LL> pii;
 38 typedef pair<LL, pii> tr;
 39 tr operator + (tr A, tr B) {
 40     return {A.first * B.first % k, {max(A.second.first, B.second.first), min(A.second.second, B.second.second)}};
 41 }
 42 tr query(int p, int tl, int tr, int l, int r) {
 43     if (tl > tr) return {1, {-INF, INF}};
 44     if (tr < l || r < tl) return {1, {-INF, INF}};
 45     if (l <= tl && tr <= r) return {pr[p], {ma[p], mi[p]}};
 46     int mid = (tl + tr) >> 1;
 47     return query(p << 1, tl, mid, l, r) + query(p << 1 | 1, mid + 1, tr, l, r);
 48 }
 49 
 50 tr ret[maxm];
 51 bool cmp1(int i, int j) {return xl[i] > xl[j];}
 52 bool cmp2(int i, int j) {return xr[i] < xr[j];}
 53 void solve(int l, int r, vector<int> Q) {
 54     int mid = (l + r) >> 1, p;
 55     vector<int> QL, QM, QR;
 56     for(int i = 0; i < Q.size(); ++i) {
 57         int o = Q[i];
 58         if(l != r && xr[o] <= mid) QL.push_back(o);
 59         else if(l != r && xl[o] > mid) QR.push_back(o);
 60         else QM.push_back(o);
 61     }
 62     sort(QM.begin(), QM.end(), cmp1);
 63     p = mid + 1;
 64     for(int i = 0; i < QM.size(); ++i) {
 65         int o = QM[i];
 66         while(p > xl[o]) --p, modify(1, 1, n, y[p], w[p]);
 67         ret[o] = query(1, 1, n, yl[o], yr[o]);
 68     }
 69     while(p <= mid) modify(1, 1, n, y[p], 0), p++;
 70     p--;
 71     sort(QM.begin(), QM.end(), cmp2);
 72     for(int i = 0; i < QM.size(); ++i) {
 73         int o = QM[i];
 74         while(p < xr[o]) ++p, modify(1, 1, n, y[p], w[p]);
 75         ret[o] = ret[o] + query(1, 1, n, yl[o], yr[o]);
 76         if(ret[o].second.first == -INF) P[o] = MAX[o] = MIN[o] = 0;
 77         else P[o] = ret[o].first, MAX[o] = ret[o].second.first, MIN[o] = ret[o].second.second;
 78     }
 79     while(p > mid) modify(1, 1, n, y[p], 0), p--;
 80     if(!QL.empty()) solve(1, mid, QL);
 81     if(!QR.empty()) solve(mid + 1, r, QR);
 82 }
 83 
 84 int main() {
 85     int T;
 86     scanf("%d", &T);
 87     while(T--) {
 88         scanf("%d", &n);
 89         for(int i = 1; i <= n; ++i) scanf("%lld %lld", y + i, w + i);
 90         LL m, a0, b0, c0, d0, p, q, r, MOD;
 91         scanf("%lld %lld %lld %lld %lld %lld %lld %lld %lld %lld", &m, &a0, &b0, &c0, &d0, &p, &q, &r, &MOD, &k);
 92         vector<int> Q;
 93         for(int i = 1; i <= m; ++i) {
 94             LL ai = (p * a0 + q * b0 + r) % MOD;
 95             LL bi = (p * b0 + q * a0 + r) % MOD;
 96             LL ci = (p * c0 + q * d0 + r) % MOD;
 97             LL di = (p * d0 + q * c0 + r) % MOD;
 98             a0 = ai, b0 = bi, c0 = ci, d0 = di;
 99             xl[i] = a0 % n + 1, xr[i] = b0 % n + 1;
100             if(xl[i] > xr[i]) swap(xl[i], xr[i]);
101             yl[i] = c0 % n + 1, yr[i] = d0 % n + 1;
102             if(yl[i] > yr[i]) swap(yl[i], yr[i]);
103             Q.push_back(i);
104         }
105         build(1, 1, n), solve(1, n, Q);
106         LL ans = 0;
107         for(int i = 1; i <= m; ++i) ans = ans + (P[i] ^ MAX[i] ^ MIN[i]);
108         printf("%lld
", ans);
109     }
110     return 0;
111 }
    Aguin

    Problem K. Transport Construction

    那个牛逼哄哄的Boruvka？啥玩意儿好像就是每次对每个连通块找到边权最小的边，把这些边合并一下下，这样每次连通块至少减少一半，所以只有logn次

    然后点乘最小就是拿一个垂线从下往上扫一下，最先碰到的就是投影最小的，所以一定在下凸壳上

    然后再每次按连通块编号快乐分治，下凸壳里面是斜率递增的，所以询问的向量也按斜率递增排，双指针搞搞，好像也没有啥要注意的……

      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 LL X[maxn], Y[maxn];
  7 LL cross(int i, int j, int k) {return (X[i] - X[k]) * (Y[j] - Y[k]) - (Y[i] - Y[k]) * (X[j] - X[k]);}
  8 LL dot(int i, int j) {return X[i] * X[j] + Y[i] * Y[j];}
  9 
 10 // UF
 11 int fa[maxn];
 12 int Find(int x) {
 13     return fa[x] == x ? x : fa[x] = Find(fa[x]);
 14 }
 15 void Union(int x, int y) {
 16     fa[Find(x)] = Find(y);
 17 }
 18 
 19 // Tarjan
 20 stack<int> S;
 21 vector<int> G[maxn], bcc[maxn];
 22 int dfs_clock, dfn[maxn], low[maxn];
 23 int bcc_cnt, bccno[maxn];
 24 void dfs(int u) {
 25     dfn[u] = low[u] = ++dfs_clock;
 26     S.push(u);
 27     for (int i = 0; i < G[u].size(); i++) {
 28         int v = G[u][i];
 29         if (!dfn[v]) {
 30             dfs(v);
 31             low[u] = min(low[u], low[v]);
 32         } else if (!bccno[v]) low[u] = min(low[u], dfn[v]);
 33     }
 34     if (low[u] == dfn[u]) {
 35         bcc_cnt++;
 36         while (1) {
 37             int x = S.top();
 38             S.pop();
 39             bccno[x] = bcc_cnt;
 40             bcc[bcc_cnt].push_back(x);
 41             if (x == u) break;
 42         }
 43     }
 44 }
 45 void find_bcc(int n) {
 46     dfs_clock = bcc_cnt = 0;
 47     for (int i = 1; i <= n; ++i) dfn[i] = bccno[i] = 0, bcc[i].clear();
 48     for (int i = 1; i <= n; i++) if (!dfn[i]) dfs(i);
 49 }
 50 
 51 // D & C
 52 LL M[maxn];
 53 pii pr[maxn];
 54 bool cmpx(int i, int j) {return X[i] < X[j];}
 55 bool cmpk(int i, int j) {return Y[i] * X[j] < Y[j] * X[i];}
 56 vector<int> hull[maxn << 2], line[maxn << 2];
 57 void solve(int p, int l, int r) {
 58     hull[p].clear(), line[p].clear();
 59     if(l == r) {
 60         line[p] = bcc[l];
 61         sort(line[p].begin(), line[p].end(), cmpx);
 62         for(int i = 0; i < line[p].size(); ++i) {
 63             int x = line[p][i], sz = hull[p].size();
 64             while(sz > 1 && cross(hull[p][sz - 1], x, hull[p][sz - 2]) <= 0) hull[p].pop_back(), sz--;
 65             hull[p].push_back(x);
 66         }
 67         sort(line[p].begin(), line[p].end(), cmpk);
 68         return;
 69     }
 70     int mid = (l + r) / 2, p1 = 0, p2 = 0;
 71     solve(p << 1, l, mid), solve(p << 1 | 1, mid + 1, r);
 72     for(int i = 0; i < line[p << 1].size(); ++i) {
 73         int x = line[p << 1][i], bx = bccno[x];
 74         while(p1 + 1 < hull[p << 1 | 1].size() && dot(x, hull[p << 1 | 1][p1 + 1]) <= dot(x, hull[p << 1 | 1][p1])) p1++;
 75         int y = hull[p << 1 | 1][p1];
 76         LL DOT = dot(x, y);
 77         if(DOT <= M[bx]) M[bx] = DOT, pr[bx] = pii(x, y);
 78     }
 79     for(int i = 0; i < line[p << 1 | 1].size(); ++i) {
 80         int x = line[p << 1 | 1][i], bx = bccno[x];
 81         while(p2 + 1 < hull[p << 1].size() && dot(x, hull[p << 1][p2 + 1]) <= dot(x, hull[p << 1][p2])) p2++;
 82         int y = hull[p << 1][p2];
 83         LL DOT = dot(x, y);
 84         if(DOT <= M[bx]) M[bx] = DOT, pr[bx] = pii(x, y);
 85     }
 86     p1 = p2 = 0;
 87     while(p1 < line[p << 1].size() || p2 < line[p << 1 | 1].size()) {
 88         if(p1 == line[p << 1].size()) line[p].push_back(line[p << 1 | 1][p2++]);
 89         else if(p2 == line[p << 1 | 1].size() || cmpk(line[p << 1][p1], line[p << 1 | 1][p2])) line[p].push_back(line[p << 1][p1++]);
 90         else line[p].push_back(line[p << 1 | 1][p2++]);
 91     }
 92     p1 = p2 = 0;
 93     while(p1 < hull[p << 1].size() || p2 < hull[p << 1 | 1].size()) {
 94         int x, sz = hull[p].size();
 95         if(p1 == hull[p << 1].size()) x = hull[p << 1 | 1][p2++];
 96         else if(p2 == hull[p << 1 | 1].size() || cmpx(hull[p << 1][p1], hull[p << 1 | 1][p2])) x = hull[p << 1][p1++];
 97         else x = hull[p << 1 | 1][p2++];
 98         while(sz > 1 && cross(hull[p][sz - 1], x, hull[p][sz - 2]) <= 0) hull[p].pop_back(), sz--;
 99         hull[p].push_back(x);
100     }
101 }
102 
103 int main() {
104     int T;
105     scanf("%d", &T);
106     while(T--) {
107         LL ans = 0;
108         int n, cnt = 0;
109         scanf("%d", &n);
110         for(int i = 1; i <= n; ++i) scanf("%lld %lld", X + i, Y + i), fa[i] = i, G[i].clear();
111         while(cnt < n - 1) {
112             find_bcc(n);
113             for(int i = 1; i <= bcc_cnt; ++i) M[i] = 1e18;
114             solve(1, 1, bcc_cnt);
115             for(int i = 1; i <= bcc_cnt; ++i) {
116                 int x = pr[i].first, y = pr[i].second;
117                 if(Find(x) == Find(y)) continue;
118                 Union(x, y), ans += M[i], cnt++;
119                 G[x].push_back(y), G[y].push_back(x);
120             }
121         }
122         printf("%lld
", ans);
123     }
124     return 0;
125 }
    Aguin

    Problem L. Visual Cube

    甩锅给谈学姐他竟然写不出来

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 void printChar(int n, char c) {
 5     if (n <= 0) return;
 6     while (n--) putchar(c);
 7     return;
 8 }
 9 
10 int main () {
11     int a, b, c, N;
12     scanf("%d", &N);
13     while (N--) {
14         scanf("%d%d%d", &a, &b, &c);
15         int pnts;
16         for (int i = 0; i < 2 * c + 2 * b + 1; i++) {
17             printChar(2 * b - i, '.');
18             for (int j = 0; j < 2 * a + 1; j++) {
19                 if (i % 2)
20                     putchar((j%2)?'.':((2 * b > i)?'/':'|'));
21                 else
22                     putchar((j%2)?'-':'+');
23             }
24             pnts = 2 * a;
25             if (2 * b > i) pnts += 2 * b - i;
26             for (int j = pnts; j < 2 * a + 2 * b; j++) {
27                 if (i >= 2 * c + 1 && j >= 2 * (b + a + c) - i)
28                     putchar('.');
29                 else
30                     if (i % 2)
31                         putchar((j%2)?'|':'/');
32                     else
33                         putchar((j%2)?'+':'.');
34             }
35             puts("");
36         }
37     }
38 }
    摸瑜瑜

    Problem M. Walking Plan

    其实应该先做一遍Floyd……

     1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL INF = 1e18;
 5 LL a[101][55][55], b[101][55][55];
 6 
 7 int main() {
 8     int T;
 9     scanf("%d", &T);
10     while(T--) {
11         int n, m, q;
12         scanf("%d %d", &n, &m);
13         for(int i = 1; i <= n; ++i)
14             for(int j = 1; j <= n; ++j)
15                 b[1][i][j] = INF;
16         for(int i = 1; i <= m; ++i) {
17             int u, v, w;
18             scanf("%d %d %d", &u, &v, &w);
19             b[1][u][v] = min(b[1][u][v], (LL) w);
20         }
21         for(int j = 1; j <= n; ++j)
22             for(int k = 1; k <= n; ++k)
23                 for(int p = 1; p <= n; ++p)
24                     b[1][k][p] = min(b[1][k][p], b[1][k][j] + b[1][j][p]);
25         for(int i = 2; i <= 100; ++i) {
26             for(int j = 1; j <= n; ++j)
27                 for(int k = 1; k <= n; ++k)
28                     b[i][j][k] = INF;
29             for(int j = 1; j <= n; ++j)
30                 for(int k = 1; k <= n; ++k)
31                     for(int p = 1; p <= n; ++p)
32                         b[i][k][p] = min(b[i][k][p], b[i - 1][k][j] + b[1][j][p]);
33         }
34         for(int j = 1; j <= n; ++j)
35             for(int k = 1; k <= n; ++k)
36                 a[1][j][k] = b[100][j][k];
37         for(int i = 2; i <= 100; ++i) {
38             for(int j = 1; j <= n; ++j)
39                 for(int k = 1; k <= n; ++k)
40                     a[i][j][k] = INF;
41             for(int j = 1; j <= n; ++j)
42                 for(int k = 1; k <= n; ++k)
43                     for(int p = 1; p <= n; ++p)
44                         a[i][k][p] = min(a[i][k][p], a[i - 1][k][j] + a[1][j][p]);
45         }
46         scanf("%d", &q);
47         while(q--) {
48             int s, t, k;
49             scanf("%d %d %d", &s, &t, &k);
50             int A = (k - 1) / 100, B = k % 100 == 0 ? 100 : k % 100;
51             LL ans = INF;
52             if(A) for(int i = 1; i <= n; ++i) ans = min(ans, a[A][s][i] + b[B][i][t]);
53             else ans = min(ans, b[B][s][t]);
54             printf("%lld
", ans == INF ? -1 : ans);
55         }
56     }
57     return 0;
58 }
    Aguin
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  • 原文地址：https://www.cnblogs.com/Aguin/p/9399842.html
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