2016 Multi-University Training Contest 1

先做最小生成树（边权不同保证唯一）。

然后打牌一下sz，单边贡献为边权乘两头sz积。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <vector>
  5 #include <algorithm>
  6 using namespace std;
  7 typedef long long LL;
  8 const int maxn = 1e5 + 10;
  9 int n, m;
 10 LL ans;
 11 
 12 
 13 // G
 14 struct Edge
 15 {
 16     int from, to, val;
 17     Edge(int F, int T, int V): from(F), to(T), val(V){}
 18     friend bool operator < (Edge A, Edge B)
 19     {
 20         return A.val < B.val;
 21     }
 22 };
 23 vector<Edge> vec;
 24 
 25 
 26 // UF
 27 int fa[maxn];
 28 int Find(int x)
 29 {
 30     return fa[x] == x ? x : fa[x] = Find(fa[x]);
 31 }
 32 void Union(int x, int y)
 33 {
 34     x = Find(x);
 35     y = Find(y);
 36     fa[y] = x;
 37 }
 38 
 39 
 40 // tree
 41 int cnt, h[maxn];
 42 struct edge
 43 {
 44     int to, pre, val;
 45 } e[maxn<<1];
 46 void add(int from, int to, int val)
 47 {
 48     cnt++;
 49     e[cnt].pre = h[from];
 50     e[cnt].to = to;
 51     e[cnt].val = val;
 52     h[from] = cnt;
 53 }
 54 
 55 
 56 // Kruskal
 57 LL Kruskal()
 58 {
 59     LL ret = 0LL;
 60     sort(vec.begin(), vec.end());
 61     int sz = vec.size();
 62     for(int i = 0; i < sz; i++)
 63     {
 64         Edge tmp = vec[i];
 65         int a = tmp.from, b = tmp.to, v = tmp.val;
 66         if(Find(a) == Find(b)) continue;
 67         add(a, b, v), add(b, a, v);
 68         Union(a, b);
 69         ret = ret + v;
 70     }
 71     return ret;
 72 }
 73 
 74 
 75 // dp
 76 int sz[maxn];
 77 void dfs(int x, int f)
 78 {
 79     sz[x] = 1;
 80     for(int i = h[x]; i; i = e[i].pre)
 81     {
 82         int to = e[i].to, v = e[i].val;
 83         if(to == f) continue;
 84         dfs(to, x);
 85         sz[x] += sz[to];
 86         ans += (LL) v * sz[to] * (n - sz[to]);
 87     }
 88 }
 89 
 90 
 91 void init()
 92 {
 93     ans = 0LL;
 94     cnt = 0;
 95     vec.clear();
 96     memset(h, 0, sizeof(h));
 97     for(int i = 0; i < maxn; i++) fa[i] = i;
 98 }
 99 
100 
101 int main(void)
102 {
103     int T;
104     scanf("%d", &T);
105     while(T--)
106     {
107         init();
108         scanf("%d %d", &n, &m);
109         for(int i = 1; i <= m; i++)
110         {
111             int a, b, v;
112             scanf("%d %d %d", &a, &b, &v);
113             vec.push_back(Edge(a, b, v));
114         }
115         LL ans1 = Kruskal();
116         dfs(1, 0);
117         printf("%I64d %.2f\n", ans1, 2.0 * ans / n / (n - 1));
118     }
119     return 0;
120 }

Aguin

1002 Chess

单行2^20局面暴力打好sg，每行^一下。

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int sg[1<<20];
 5 
 6 void cal(int x)
 7 {
 8     int last = -1, vis[20] = {0};
 9     for(int i = 0; i < 20; i++)
10     {
11         if(!(x & (1 << i))) last = i;
12         else if(last != -1) vis[sg[((1<<last)|(1<<i))^x]] = 1;
13     }
14     for(int i = 0; i <= 20; i++)
15     if(!vis[i]) {sg[x] = i; return;}
16 }
17 
18 int main(void)
19 {
20     for(int i = 0; i < (1 << 20); i++) cal(i);
21 
22     int T;
23     scanf("%d", &T);
24     while(T--)
25     {
26         int N, ans = 0;
27         scanf("%d", &N);
28         for(int i = 1; i <= N; i++)
29         {
30             int m, x = 0;
31             scanf("%d", &m);
32             for(int i = 0; i < m; i++)
33             {
34                 int y;
35                 scanf("%d", &y);
36                 x += 1 << (20 - y);
37             }
38             ans ^= sg[x];
39         }
40         puts(ans ? "YES" : "NO");
41     }
42     return 0;
43 }

Aguin

1003 Game

1004 GCD

打好st表，对每个固定的左端点，gcd最多只有logai个下降点。

可以二分每个下降点的位置，更好的方法是从右往左枚举左端点，维护下降点序列。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <map>
 4 #include <set>
 5 using namespace std;
 6 typedef long long LL;
 7 const int maxn = 1e5 + 10;
 8 map<LL, LL> M;
 9 set<int> S;
10 set<int> :: iterator it;
11 int N;
12 
13 LL gcd(LL a, LL b)
14 {
15     return a % b ? gcd(b, a % b) : b;
16 }
17 
18 // RMQ
19 LL a[maxn], rmq[maxn][20];
20 void RMQ_init()
21 {
22     for(int i = 1; i <= N; i++) rmq[i][0] = a[i];
23     for(int j = 1; (1 << j) <= N; j++)
24         for(int i = 1; i + ( 1 << j ) - 1 <= N; i++)
25             rmq[i][j] = gcd(rmq[i][j-1] , rmq[i+(1<<j-1)][j-1]);
26 }
27 int RMQ_query(int l, int r)
28 {
29     int k = 0;
30     while( ( 1 << (k + 1) ) <= r - l + 1 ) k++;
31     return gcd(rmq[l][k], rmq[r-(1<<k)+1][k]);
32 }
33 
34 int main(void)
35 {
36     int T;
37     scanf("%d", &T);
38     for(int kase = 1; kase <= T; kase++)
39     {
40         scanf("%d", &N);
41         for(int i = 1; i <= N; i++) scanf("%I64d", a + i);
42         RMQ_init();
43 
44         S.clear(), M.clear();
45         for(int l = N; l; l--)
46         {
47             int last = N;
48             LL now = RMQ_query(l, N);
49             for(it = S.begin(); it != S.end();)
50             {
51                 LL tmp = RMQ_query(l, -(*it));
52                 if(tmp == now) S.erase(it++);
53                 else
54                 {
55                     M[now] += (LL) (last + *it);
56                     now = tmp;
57                     last = -(*(it++));
58                 }
59             }
60             M[now] += (LL) (last - l);
61             M[a[l]]++;
62             if(a[l] != now) S.insert(-l);
63         }
64 
65         int Q;
66         scanf("%d", &Q);
67         printf("Case #%d:\n", kase);
68         for(int i = 0; i < Q; i++)
69         {
70             int l, r;
71             scanf("%d %d", &l, &r);
72             LL ans = RMQ_query(l, r);
73             printf("%I64d %I64d\n", ans, M[ans]);
74         }
75 
76     }
77     return 0;
78 }

Aguin

1005 Necklace

枚举yin排列，以yang和空位建二分图。

空位两边yin不会使yang褪色则连边，N - 最大匹配即为答案。

习惯用Dinic跑，结果1s+。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <queue>
  6 using namespace std;
  7 const int INF = 1e9;
  8 int lv[22], it[22];
  9 int cnt, h[22];
 10 
 11 struct edge
 12 {
 13     int to, pre, cap;
 14 } e[1111];
 15 
 16 void init()
 17 {
 18     memset(h, -1, sizeof(h));
 19     cnt = 0;
 20 }
 21 
 22 void add(int from, int to, int cap)
 23 {
 24     e[cnt].pre = h[from];
 25     e[cnt].to = to;
 26     e[cnt].cap = cap;
 27     h[from] = cnt;
 28     cnt++;
 29 }
 30 
 31 void ad(int from, int to, int cap)
 32 {
 33     add(from, to, cap);
 34     add(to, from, 0);
 35 }
 36 
 37 void bfs(int s)
 38 {
 39     memset(lv, -1, sizeof(lv));
 40     queue<int> q;
 41     lv[s] = 0;
 42     q.push(s);
 43     while(!q.empty())
 44     {
 45         int v = q.front(); q.pop();
 46         for(int i = h[v]; i >= 0; i = e[i].pre)
 47         {
 48             int cap = e[i].cap, to = e[i].to;
 49             if(cap > 0 && lv[to] < 0)
 50             {
 51                 lv[to] = lv[v] + 1;
 52                 q.push(to);
 53             }
 54         }
 55     }
 56 }
 57 
 58 int dfs(int v, int t, int f)
 59 {
 60     if(v == t) return f;
 61     for(int &i = it[v]; i >= 0; i = e[i].pre)
 62     {
 63         int &cap = e[i].cap, to = e[i].to;
 64         if(cap > 0 && lv[v] < lv[to])
 65         {
 66             int d = dfs(to, t, min(f, cap));
 67             if(d > 0)
 68             {
 69                 cap -= d;
 70                 e[i^1].cap += d;
 71                 return d;
 72             }
 73         }
 74     }
 75     return 0;
 76 }
 77 
 78 int Dinic(int s, int t)
 79 {
 80     int flow = 0;
 81     while(1)
 82     {
 83         bfs(s);
 84         if(lv[t] < 0) return flow;
 85         memcpy(it, h, sizeof(it));
 86         int f;
 87         while((f = dfs(s, t, INF)) > 0) flow += f;
 88     }
 89 }
 90 
 91 int a[11], G[11][11];
 92 int main(void)
 93 {
 94     int N, M;
 95     while(~scanf("%d %d", &N, &M))
 96     {
 97         if(!N) {puts("0"); continue;}
 98         memset(G, 0, sizeof(G));
 99         for(int i = 1; i <= M; i++)
100         {
101             int a, b;
102             scanf("%d %d", &a, &b);
103             G[a][b] = 1;
104         }
105 
106         int ans = 0;
107         for(int i = 0; i < N; i++) a[i] = i + 1;
108         while(1)
109         {
110             init();
111             int S = N + N + 1, T = S + 1;
112             for(int i = 1; i <= N; i++) ad(S, i, 1), ad(N + i, T, 1);
113             for(int i = 1; i <= N; i++)
114                 for(int j = 0; j < N; j++)
115                     if(!G[i][a[j]] && !G[i][a[(j+1)%N]])
116                         ad(i, N + 1 + j, 1);
117             ans = max(ans, Dinic(S, T));
118             if(!next_permutation(a + 1, a + N)) break;
119         }
120         printf("%d\n", N - ans);
121     }
122     return 0;
123 }

Aguin

换匈牙利倒是快些。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 int N, cnt, h[11];
 7 
 8 struct edge
 9 {
10     int to, pre;
11 } e[1111];
12 
13 void init()
14 {
15     memset(h, -1, sizeof(h));
16     cnt = 0;
17 }
18 
19 void add(int from, int to)
20 {
21     e[cnt].pre = h[from];
22     e[cnt].to = to;
23     h[from] = cnt;
24     cnt++;
25 }
26 
27 // Hungary
28 int lk[11], vis[11];
29 bool dfs(int x)
30 {
31     for(int i = h[x]; i != -1; i = e[i].pre)
32     {
33         int to = e[i].to;
34         if(!vis[to])
35         {
36             vis[to] = 1;
37             if(lk[to] == -1 || dfs(lk[to]))
38             {
39                 lk[to] = x;
40                 return true;
41             }
42         }
43     }
44     return false;
45 }
46 int Hungary()
47 {
48     int ret = 0;
49     memset(lk, -1, sizeof(lk));
50     for(int i = 1; i <= N; i++)
51     {
52         memset(vis, 0, sizeof(vis));
53         if(dfs(i)) ret++;
54     }
55     return ret;
56 }
57 
58 
59 int a[11], G[11][11];
60 int main(void)
61 {
62     int M;
63     while(~scanf("%d %d", &N, &M))
64     {
65         if(!N) {puts("0"); continue;}
66         memset(G, 0, sizeof(G));
67         for(int i = 1; i <= M; i++)
68         {
69             int a, b;
70             scanf("%d %d", &a, &b);
71             G[a][b] = 1;
72         }
73 
74         int ans = 0;
75         for(int i = 0; i < N; i++) a[i] = i + 1;
76         while(1)
77         {
78             init();
79             for(int i = 1; i <= N; i++)
80                 for(int j = 0; j < N; j++)
81                     if(!G[i][a[j]] && !G[i][a[(j+1)%N]])
82                         add(i, 1 + j);
83             ans = max(ans, Hungary());
84             if(!next_permutation(a + 1, a + N)) break;
85         }
86         printf("%d\n", N - ans);
87     }
88     return 0;
89 }

Aguin

1006 PowMod

1007 Rigid Frameworks

首先问题等价于求连通二分图数目。

据鸟神所说，m行n列的桁架，在第i行第j列加约束，使得：

①第i行的所有竖向杆件平行。

②第j列的所有横向杆件平行。

③第i行所有竖向杆件和第j列所有横向杆件垂直。

故只要对左边m个点，右边n个点建二分图，

只要二分图连通，则所有横向杆件平行，所有竖向杆件平行，所有横向杆件与所有纵向杆件垂直，即为刚体。

连通二分图计数问题，可以照着一般的连通图计数画葫芦。

参考：gyz营员交流

令$S(n, m)$为左边n个点，右边m个点的二分图数。

$F(n, m)$为左边n个点，右边m个点的连通二分图数。

$G(n, m) = S(n, m) - F(n, m)$

固定二分图左边的第一个点，考虑有多少个点与其连通，

$G(n, m) = \sum_{i = 0}^{n - 1} \sum_{j = 0, i + j \neq m + n - 1}^{m} C_{n-1}^{i}C_{m}^{j} F(i+1, j) S(n-i-1, m-j)$

而本题中对于i行j列的一个格子，可以选择不加约束，加左斜杠，加右斜杠。

故$S(n, m) = 3 ^ {nm}$

最后需要注意的是$F(n, m)$的边界条件。

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 typedef long long LL;
 5 const LL mod = 1e9 + 7;
 6 LL S[11][11], F[11][11], G[11][11];
 7 LL c[11][11], p[111];
 8 
 9 int main(void)
10 {
11     p[0] = 1LL;
12     for(int i = 1; i <= 100; i++) p[i] = p[i-1] * 3 % mod;
13 
14     for(int i = 0; i <= 10; i++) c[i][0] = c[i][i] = 1;
15     for(int i = 1; i <= 10; i++)
16         for(int j = 1; j < i; j++)
17             c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod;
18 
19     F[0][1] = F[1][0] = 1;
20     for(int i = 0; i <= 10; i++) S[0][i] = S[i][0] = 1;
21     for(int i = 1; i <= 10; i++)
22     for(int j = 1; j <= 10; j++)
23     {
24         for(int p = 0; p <= i - 1; p++)
25         for(int q = 0; q <= j && p + q != i + j - 1; q++)
26         G[i][j] = (G[i][j] + c[i-1][p] * c[j][q] % mod * F[p+1][q] % mod * S[i-p-1][j-q] % mod) % mod;
27 
28         S[i][j] = p[i*j];
29         F[i][j] = (S[i][j] - G[i][j] + mod) % mod;
30     }
31 
32 
33     int m, n;
34     while(~scanf("%d %d", &m, &n)) printf("%I64d\n", F[m][n]);
35     return 0;
36 }

Aguin

1008 Shell Necklace

1009 Solid Dominoes Tilings

1010 Subway

树同构，哈希一下就可以了。

题解的哈希方法以前看csy说过。

大概就是先找重心，每个点的孩子按哈希值排序算哈希值。

然而一开始两个素数随便取老冲突，于是以sz作为第二关键字排，就稳了。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <vector>
  5 #include <algorithm>
  6 #include <map>
  7 using namespace std;
  8 typedef long long LL;
  9 const int maxn = 2e5 + 10;
 10 const LL P = 1e6 + 7, Q = 1e9 + 7;
 11 int N;
 12 
 13 map<string, int> M1, M2;
 14 string name[maxn];
 15 
 16 // edge
 17 int cnt, h[maxn];
 18 void init()
 19 {
 20     cnt = 0;
 21     memset(h, 0, sizeof(h));
 22 }
 23 struct edge
 24 {
 25     int to, pre;
 26 } e[maxn<<1];
 27 void add(int from, int to)
 28 {
 29     cnt++;
 30     e[cnt].pre = h[from];
 31     e[cnt].to = to;
 32     h[from] = cnt;
 33 }
 34 
 35 
 36 // Node
 37 struct node
 38 {
 39     int id, sz;
 40     LL h;
 41     node(int I, LL H, int S): id(I), h(H), sz(S) {}
 42     friend bool operator < (node A, node B)
 43     {
 44         if(A.h != B.h) return A.h < B.h;
 45         return A.sz < B.sz;
 46     }
 47 };
 48 
 49 
 50 // dfs
 51 int sz[maxn], M;
 52 vector<int> G;
 53 void dfs1(int x, int f)
 54 {
 55     sz[x] = 1;
 56     int tmp = 0;
 57     for(int i = h[x]; i; i = e[i].pre)
 58     {
 59         int to = e[i].to;
 60         if(to == f) continue;
 61         dfs1(to, x);
 62         sz[x] += sz[to];
 63         tmp = max(tmp, sz[to]);
 64     }
 65     tmp = max(tmp, N - sz[x]);
 66     if(tmp < M)
 67     {
 68         M = tmp;
 69         G.clear();
 70         G.push_back(x);
 71     }
 72     else if(tmp == M) G.push_back(x);
 73 }
 74 
 75 
 76 LL H[maxn];
 77 vector<node> v[maxn];
 78 void dfs2(int x, int f)
 79 {
 80     sz[x] = 1;
 81     v[x].clear();
 82     for(int i = h[x]; i; i = e[i].pre)
 83     {
 84         int to = e[i].to;
 85         if(to == f) continue;
 86         dfs2(to, x);
 87         sz[x] += sz[to];
 88         v[x].push_back(node(to, H[to], sz[to]));
 89     }
 90 
 91     sort(v[x].begin(), v[x].end());
 92     H[x] = 0;
 93     int sv = v[x].size();
 94     LL c = P;
 95     if(!sv) H[x] = 1;
 96     for(int i = 0; i < sv; i++)
 97     {
 98         H[x] = (H[x] + v[x][i].h * c) % Q;
 99         c = c * P % Q;
100     }
101 }
102 
103 void dfs3(int x1, int x2)
104 {
105     cout << name[x1] << ' ' << name[x2] << endl;
106     int sv = v[x1].size();
107     for(int i = 0; i < sv; i++) dfs3(v[x1][i].id, v[x2][i].id);
108 }
109 
110 int main(void)
111 {
112 
113     ios::sync_with_stdio(0);
114 
115     while(cin >> N)
116     {
117         init();
118         M1.clear(), M2.clear();
119         int num = 0;
120 
121         for(int i = 1; i < N; i++)
122         {
123             string a, b;
124             cin >> a >> b;
125             if(!M1[a]) M1[a] = ++num, name[num] = a;
126             if(!M1[b]) M1[b] = ++num, name[num] = b;
127             add(M1[a], M1[b]), add(M1[b], M1[a]);
128         }
129 
130         M = N + 1;
131         dfs1(1, 0);
132         dfs2(G[0], 0);
133         int rt1 = G[0];
134 
135         for(int i = 1; i < N; i++)
136         {
137             string a, b;
138             cin >> a >> b;
139             if(!M2[a]) M2[a] = ++num, name[num] = a;
140             if(!M2[b]) M2[b] = ++num, name[num] = b;
141             add(M2[a], M2[b]), add(M2[b], M2[a]);
142         }
143 
144         M = N + 1;
145         dfs1(N + 1, 0);
146         dfs2(G[0], 0);
147         int rt2 = G[0];
148         if(H[rt1] != H[G[0]]) dfs2(G[1], 0), rt2 = G[1];
149 
150         dfs3(rt1, rt2);
151 
152     }
153     return 0;
154 }

Aguin

1011 tetrahedron

posted @ 2016-07-21 21:26 Aguin 阅读(244) 评论(0) 编辑收藏举报

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2016 Multi-University Training Contest 1

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