2016 Multi-University Training Contest 1
8/11
2016 Multi-University Training Contest 1
最小生成树+线性期望 A Abandoned country(BH)
题意:
1. 求最小生成树 2. 求在某一棵最小生成树任意两点的最小距离的期望值。
思路:
首先题目说了边权值都是不同的,所以最小生成树唯一。那么只要统计出最小生成树的每一条边在“任意两点走经过它“的情况下所贡献的值,发现在一棵树里,一条边所贡献的次数为,sz[v]表示v子树包括节点v的个数。如下图所示,红边所贡献的次数即为”选一个蓝点再选一个绿点“的方案数。
代码:
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 1e5 + 5; const int M = 1e6 + 5; struct Edge { int u, v, w; bool operator < (const Edge &rhs) const { return w < rhs.w; } }edges[M]; int n, m; vector<pair<int, int> > edge[N]; void init_edge() { for (int i=1; i<=n; ++i) { edge[i].clear (); } } int rt[N]; void init_DSU() { for (int i=1; i<=n; ++i) { rt[i] = i; } } int Find(int x) { return rt[x] == x ? x : rt[x] = Find (rt[x]); } int sz[N]; ll sumw; void DFS2(int u, int fa) { for (auto t: edge[u]) { int v = t.first, w = t.second; if (v == fa) continue; sumw += (ll) w * (n - sz[v]) * sz[v]; DFS2 (v, u); } } void DFS(int u, int fa) { sz[u] = 1; for (auto t: edge[u]) { int v = t.first, w = t.second; if (v == fa) continue; DFS (v, u); sz[u] += sz[v]; } } double solve() { memset (sz, 0, sizeof (sz)); sumw = 0; DFS (1, 0); DFS2 (1, 0); return ((double) sumw * 2 / n / (n - 1)); } int main() { int T; scanf ("%d", &T); while (T--) { scanf ("%d%d", &n, &m); init_DSU (); int tot = 0; for (int i=1; i<=m; ++i) { int u, v, w; scanf ("%d%d%d", &u, &v, &w); edges[tot++] = (Edge) {u, v, w}; } std::sort (edges, edges+tot); init_edge (); ll sum = 0; for (int i=0; i<tot; ++i) { int u = edges[i].u, v = edges[i].v, w = edges[i].w; int fu = Find (u), fv = Find (v); if (fu == fv) continue; sum += w; rt[fv] = fu; edge[u].push_back ({v, w}); edge[v].push_back ({u, w}); } printf ("%I64d %.2f\n", sum, solve ()); } return 0; }
状态压缩+博弈 B Chess(BH)
题意:
有n*20的格子,某些格子上有棋子,A和B轮流操作,可以一个棋子放到右边第一个空位置,不能操作者输,A先操作,问输赢。
思路:
这是简单的Nim问题,状态压缩,算出所有状态的sg值。看完训练指南的内容再结合代码应该能懂了吧。
代码:
#include <bits/stdc++.h> int a[1005]; int sg[(1<<20)+5]; int vis[25]; int mex(int u) { memset (vis, 0, sizeof (vis)); for (int i=0; i<20; ++i) { if (u & (1<<i)) { int j = i - 1; while (j >= 0 && (u & (1<<j))) j--; if (j >= 0) { int v = u ^ (1<<i) ^ (1<<j); vis[sg[v]] = 1; } } } for (int i=0; ; ++i) { if (!vis[i]) return i; } } void init() { memset (sg, -1, sizeof (sg)); int x = 0; sg[0] = 0; //预处理对于所有必输态赋值为0 for (int i=0; i<20; ++i) { x |= (1<<i); sg[x] = 0; } for (int i=1; i<(1<<20); ++i) { if (sg[i] == -1) sg[i] = mex (i); } } int main() { init (); int T; scanf ("%d", &T); while (T--) { int n; scanf ("%d", &n); int ans = 0; for (int i=1; i<=n; ++i) { int m; scanf ("%d", &m); int tmp = 0; for (int j=1; j<=m; ++j) { int x; scanf ("%d", &x); //为了从小到大递推,和计算机二进制数方向相同 tmp |= (1<<(20-x)); } ans ^= sg[tmp]; } printf ("%s\n", ans > 0 ? "YES" : "NO"); } return 0; }
最短路?搜索?不会 C Game
数论(区间GCD问题)D GCD(BH)
题意:
1. 区间[l, r]的GCD值 2. 问有多少个区间的GCD值为gcd
思路:
第一个操作线段树或者ST都可以做,第二个问题的关键点是
简单来说,就是固定右端点R,然后左端点往左边移动,查询GCD,如果相同就跳到这个GCD所对应的区间的左端点(并且维护这个gcd现在的区间的两端点的位置),因为一个数的质因数最多有log(x)个,而且求GCD是递减的过程,所以跳跃的次数是log级的。
本场比赛就因为这题坑了好久,GCD的性质不够了解。学习资料
代码:
#include <bits/stdc++.h> const int N = 1e5 + 5; int a[N]; int GCD(int a, int b) { return b ? GCD (b, a % b) : a; } #define lson l, mid, o << 1 #define rson mid + 1, r, o << 1 | 1 int val[N<<2]; void push_up(int o) { val[o] = GCD (val[o<<1], val[o<<1|1]); } void build(int l, int r, int o) { if (l == r) { val[o] = a[l]; return ; } int mid = l + r >> 1; build (lson); build (rson); push_up (o); } int query(int ql, int qr, int l, int r, int o) { if (ql <= l && r <= qr) { return val[o]; } int mid = l + r >> 1, ret = 0; if (ql <= mid) ret = GCD (ret, query (ql, qr, lson)); if (qr > mid) ret = GCD (ret, query (ql, qr, rson)); return ret; } int n; std::map<int, long long> mp; int pre[N]; void init() { for (int i=1; i<=n; ++i) { pre[i] = i; } mp.clear (); } void prepare() { build (1, n, 1); for (int i=1; i<=n; ++i) { int L = i, g = a[i]; while (L > 0) { int R = L; g = GCD (g, a[L]); while (L > 1 && GCD (g, a[L-1]) == g) { L = pre[L-1]; pre[R] = L; } mp[g] += (R - L + 1); L--; } } } int main() { int T; scanf ("%d", &T); for (int cas=1; cas<=T; ++cas) { scanf ("%d", &n); init (); for (int i=1; i<=n; ++i) { scanf ("%d", &a[i]); } prepare (); int q; scanf ("%d", &q); printf ("Case #%d:\n", cas); while (q--) { int ql, qr; scanf ("%d%d", &ql, &qr); int g = query (ql, qr, 1, n, 1); printf ("%d %I64d\n", g, mp[g]); } } return 0; }
二分图匹配(匈牙利算法)E Necklace(BH)
题意:
有n颗阳属性的珠子和n颗阴属性的珠子,给了m条规则,就是某些阳珠子和阴珠子不能相邻,问2*n颗串成项链后最少几颗会违反规则。
思路:
先枚举阴珠子的排列顺序(固定第一个不动,因为是环)。然后对于某一个排列,问题可以转换为”阳珠子如何插如使得不违反规则的珠子最多“,这个用匈牙利算法解决,时间复杂度为。卿学姐DFS剪个枝也能过?数据水了吧
代码:
#include <bits/stdc++.h> int n, m; std::vector<int> edges[10]; bool no[10][10]; int lk[10]; bool vis[10]; bool DFS(int u) { for (auto v: edges[u]) { if (!vis[v]) { vis[v] = true; if (lk[v] == -1 || DFS (lk[v])) { lk[v] = u; return true; } } } return false; } int hungary() { int ret = 0; memset (lk, -1, sizeof (lk)); for (int i=1; i<=n; ++i) { memset (vis, false, sizeof (vis)); if (DFS (i)) ret++; } return ret; } int id[10]; int solve() { for (int i=1; i<=n; ++i) { id[i] = i; } int ret = 0; do { for (int i=1; i<=n; ++i) edges[i].clear (); for (int i=1; i<=n; ++i) { for (int j=1; j<=n; ++j) { if (!no[i][id[j]] && !no[i][id[(j+1>n?1:j+1)]]) { edges[i].push_back (j); } } } ret = std::max (ret, hungary ()); } while (std::next_permutation (id+2, id+1+n)); //[2,n]全排列 return ret; } int main() { while (scanf ("%d%d", &n, &m) == 2) { if (n == 0) { puts ("0"); continue; } memset (no, false, sizeof (no)); for (int i=1; i<=m; ++i) { int x, y; scanf ("%d%d", &x, &y); no[x][y] = true; } printf ("%d\n", n - solve ()); } return 0; }
数论 G PowMod(CYD)
2016多校训练一 PowMod,hdu5728(欧拉函数+指数循环节)
题意:
求 时,ans= mod p,ans的值,其中n的每个素因子的幂次都是1,是欧拉函数。
思路:
解题分两部分完成,第一部分首先求出k,有公式
(素数p是n的一个质因子)
由此可以递归求出k的值,f(n,m)=(p的欧拉值)*f(n/p,m)+f(n,m/p)。
第二部分的无限次幂,由公式
不断求欧拉值最终使值变得有限。
代码:
#include <bits/stdc++.h> using namespace std; const int M=1000000007; const int N=1e7+5; typedef long long ll; int pri[N],phi[N],tot; bool vis[N]; ll sum[N]; void init() { int n=N; tot=0; memset(vis,false,sizeof vis); phi[1]=1; for(int i=2;i<n;i++) { if(!vis[i]) { pri[tot++]=i; phi[i]=i-1; } for(int j=0;j<tot && i*pri[j]<n;j++) { vis[i*pri[j]]=true; if(i%pri[j]==0) { phi[i*pri[j]]=phi[i]*pri[j]; break; } else phi[i*pri[j]]=phi[i]*(pri[j]-1); } } sum[0]=0; for(int i=1;i<N;i++) sum[i]=(sum[i-1]+phi[i])%M; } ll Pow(ll a,ll n,ll mod) { ll ans=1; while(n) { if(n&1) { ans=ans*a%mod; } a=a*a%mod; n>>=1; } if(ans==0) ans+=mod; return ans; } ll solve(ll k,ll mod) { if(mod==1) return mod; ll tmp=phi[mod]; ll up=solve(k,tmp); ll ans=Pow(k,up,mod); return ans; } int rear; int a[15]; void resolve(ll n) { for(int i=0;i<tot;i++) { if(!vis[n]) { a[rear++]=n; break; } if(n%pri[i]==0) { a[rear++]=pri[i]; n/=pri[i]; } } } ll f(int pos,ll n,ll m) { if(n==1) return sum[m]; if(m==0) return 0; return ((a[pos]-1)*f(pos-1,n/a[pos],m)%M+f(pos,n,m/a[pos]))%M; } int main() { init(); ll n,m,p; while(scanf("%I64d%I64d%I64d",&n,&m,&p)!=EOF) { rear=0; resolve(n); ll k=f(rear-1,n,m); ll ans=solve(k,p); printf("%I64d\n",ans%p); } return 0; }
UPD:二分图连通图计数 不会 H Rigid Frameworks (BH)(51Nod原题)
题意:
给n*m的格子,可以加上斜线,问有多少种方案使得整个格子不能倾斜,也就是不能像下图一样:
思路:
网上写的不错的解题报告。问题的关键点是模型转换成二分图的连通计数,然后就是所有方案减去不符合的方案数。
代码:
#include <bits/stdc++.h> typedef long long ll; const int MOD = 1e9 + 7; ll dp[15][15][105]; ll pow_two[105]; ll C[105][105]; ll f(int a, int b, int c) { if (a >= b) return dp[a][b][c]; return dp[b][a][c]; } void init() { pow_two[0] = 1; for (int i=1; i<=100; ++i) pow_two[i] = pow_two[i-1] * 2 % MOD; C[0][0] = 1; for (int i=1; i<=100; ++i) { C[i][0] = C[i][i] = 1; for (int j=1; j<i; ++j) { C[i][j] = (C[i-1][j-1] + C[i-1][j]) % MOD; } }
//dp[n][m][k]表示n行m列取k个斜线 for (int i=1; i<=10; ++i) { dp[i][0][0] = i == 1 ? 1 : 0; for (int j=1; j<=i; ++j) { for (int k=1; k<=i*j; ++k) { dp[i][j][k] = C[i*j][k]; for (int ii=1; ii<=i; ++ii) { for (int jj=0; jj<=j; ++jj) { int kb = std::min (k, ii * jj); for (int kk=0; kk<=kb; ++kk) { if (f (ii, jj, kk) && (i-ii)*(j-jj)>=k-kk) { if (i == ii && j == jj) break; dp[i][j][k] = (dp[i][j][k] - C[i-1][ii-1] * C[j][jj] * f (ii, jj, kk) * C[(i-ii)*(j-jj)][k-kk] + MOD) % MOD; } } } } } } } } int main() { init (); int n, m; while (scanf ("%d%d", &n, &m) == 2) { if (n < m) std::swap (n, m); ll ans = 0; for (int i=1; i<=n*m; ++i) { ans = (ans + dp[n][m][i] * pow_two[i]) % MOD; } printf ("%I64d\n", ans); } return 0; }
FFT 不会 I Shell Necklace
UPD2: 轮廓线DP+容斥原理 不会 I Solid Dominoes Tilings(BH)(51Nod原题)
题意:
有n*m的格子用1*2的小格子填充,问全部填充且不能完整分割的方案数。
思路:
轮廓线DP先求全部填充的方案数(训练指南P383),再用容斥原理减去完整分割的方案数。当时觉得容斥不好做,其实容斥只要依照“奇数减,偶数加”的原则算,这是关于列的主容斥,那么对于行就把全部不符合的都减去。
代码:
#include <bits/stdc++.h> typedef long long ll; const int MOD = 1e9 + 7; const int N = 16; int dp[2][1<<N]; int f[N+1][N+1], g[N+1][N+1][1<<N], ans[N+1][N+1]; void add_mod(int &a, int b) { a += b; if (a >= MOD) a -= MOD; if (a < 0) a += MOD; } void init() { for (int n=1; n<=N; ++n) { for (int m=1; m<=N; ++m) { //预处理n*m格子填满的方案数,轮廓线DP int cur = 1; memset (dp[cur], 0, sizeof (dp[cur])); int limit = 1 << m; dp[cur][limit-1] = 1; for (int i=0; i<n; ++i) { for (int j=0; j<m; ++j) { cur ^= 1; memset (dp[cur], 0, sizeof (dp[cur])); for (int s=0; s<limit; ++s) { if (dp[cur^1][s] == 0) continue; //s状态第j位是已填 if (s & (1<<j)) { if (j && !(s&(1<<(j-1)))) { //往左放 add_mod (dp[cur][s^(1<<(j-1))], dp[cur^1][s]); } //不放 add_mod (dp[cur][s^(1<<j)], dp[cur^1][s]); } else { //往上放 add_mod (dp[cur][s^(1<<j)], dp[cur^1][s]); } } } } //保存方案数 f[n][m] = dp[cur][limit-1]; } } //容斥原理,乘法原理 std::vector<int> lens; for (int n=1; n<=N; ++n) { for (int m=1; m<=N; ++m) {
//主容斥列 //竖线最多m-1条 int limit = 1 << (m-1); for (int s=0; s<limit; ++s) { lens.clear (); int last = -1, num = 0; for (int i=0; i<m-1; ++i) { if (s & (1<<i)) { lens.push_back (i-last); last = i; num++; } } lens.push_back (m-1-last); int &now = g[n][m][s]; //f[n][len1]*f[n][len2]*... now = 1; for (int i=0; i<lens.size (); ++i) { int len = lens[i]; now = (ll) now * f[n][len] % MOD; }
//如果有横分割线,一定减去 for (int i=1; i<n; ++i) { ll up = g[i][m][s]; ll down = 1; //g[n-i][m][s] for (int j=0; j<lens.size (); ++j) { int len = lens[j]; down = (ll) down * f[n-i][len] % MOD; } add_mod (now, -(ll) up * down % MOD); } if (num & 1) add_mod (ans[n][m], -now); else add_mod (ans[n][m], now); } } } } int main() { init (); int n, m; while (scanf ("%d%d", &n, &m) == 2) { if (n < m) std::swap (n, m); printf ("%d\n", ans[n][m]); } return 0; }
树的同构判断 不会 J Subway
三维计算几何 K tetrahedron(BH,CYD)
题意:
求四面体的内切圆的坐标和半径。
思路:
求半径公式:,体积和三角形面积有函数可以算。
求圆心坐标公式:,三个坐标系分开算就可以了。
代码:
#include <bits/stdc++.h> const double EPS = 1e-10; int dcmp(double x) { return fabs (x) < EPS ? 0 : x < 0 ? -1 : 1; } struct Point3 { double x, y, z; Point3(double x=0, double y=0, double z=0) : x(x), y(y), z(z) {} }; typedef Point3 Vector3; Vector3 operator - (Point3 A, Point3 B) { return Vector3 (A.x-B.x, A.y-B.y, A.z-B.z); } //叉积 Vector3 Cross(Vector3 A, Vector3 B) { return Vector3 (A.y*B.z-A.z*B.y, A.z*B.x-A.x*B.z, A.x*B.y-A.y*B.x); } double Dot(Vector3 A, Vector3 B) { return A.x*B.x+A.y*B.y+A.z*B.z; } double Volume6(Point3 A, Point3 B, Point3 C, Point3 D) { return Dot (D-A, Cross (B-A, C-A)); } double Length(Vector3 A) { return sqrt (Dot (A, A)); } Point3 A, B, C, D; void solve() { double v6 = Volume6 (A, B, C, D); Vector3 BC = C - B, BD = D - B, BA = A - B, CA = A - C, CD = D - C; Vector3 n = Cross (BC, BD); if (dcmp (Dot (BA, n)) == 0) { puts ("O O O O"); return ; } v6 = fabs (v6); double SABC = Length (Cross (BA, BC)) * 0.5; double SBAD = Length (Cross (BA, BD)) * 0.5; double SBCD = Length (Cross (BC, BD)) * 0.5; double SCAD = Length (Cross (CD, CA)) * 0.5; double SS = SABC + SBAD + SBCD + SCAD; //V = SS * R * (1/3) double R = v6 / 2 / SS; //(1/3) Point3 O; O.x = (A.x*SBCD + B.x*SCAD + C.x*SBAD + D.x*SABC) / SS; O.y = (A.y*SBCD + B.y*SCAD + C.y*SBAD + D.y*SABC) / SS; O.z = (A.z*SBCD + B.z*SCAD + C.z*SBAD + D.z*SABC) / SS; printf ("%.4f %.4f %.4f %.4f\n", O.x, O.y, O.z, R); } int main() { while (scanf ("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &A.z, &B.x, &B.y, &B.z, &C.x, &C.y, &C.z, &D.x, &D.y, &D.z) == 12) { solve (); } return 0; }