Atcoder Dwango Programming Contest 6th 题解

(A) (Falling) (Asleep)

题目链接

(description:)

给你一个歌单(,) (n)首歌的歌名(s_i)和播放持续时间(t_i) (,)

读入一个歌名(st,)要求在歌单中算出在这首曲子之后所有曲子的播放时间总和(.)

(solution:)

暴力即可(.) 时间复杂度 (O(n + sum|s_i|))

(code:)

#include <bits/stdc++.h>
using namespace std;
inline int read(){
    int x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0',c = getchar();
    return x;
}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
inline void write(int x){if (x > 9) write(x/10); putchar(x%10+'0'); }
int n; string st,s[500]; int t[500];
int main(){
	cin >> n;
	for (int i = 1; i<= n; ++i) cin >> s[i] >> t[i];
	cin >> st; 
	int id = -1; long long ans = 0;
	for (int i = 1; i <= n; ++i) if (s[i] == st) id = i;
	for (int i = id+1; i <= n; ++i) ans += t[i];
	cout << ans << endl;
    return 0;
}

(B) (Fusing) (Slimes)

题目链接

(description :)

有(n(n≤10^5))块石子(,)每个石子有其初始坐标(x_i.)

每次你会从中随机选出不在最右边的一堆(,)将这一堆移动到 它右边最靠近它的一堆石子的位置(,)并把这两堆石子合并为一堆(.)

求出所有情况下(,)石子移动距离的总和 (,) 即期望乘以 ((n-1)!) (.)

答案对(P = 1e9 + 7) 取模(.)

(solution :)

考虑一段距离(d_i = x_{i+1} - x_i)

定义 $dp[n] = $ 一段距离之前有(n)块石子(,)把这(n)块石子移动过去之后(,)这段距离被经过的期望次数

不难得出 (dp[i] = dp[i-1] + 1/i)

然后就可以直接计算出每一段距离对答案的贡献了(.)

时间复杂度(O(n).)

(code :)

#include <bits/stdc++.h>
using namespace std;
inline int read(){
    int x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0',c = getchar();
    return x;
}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
inline void write(int x){if (x > 9) write(x/10); putchar(x%10+'0'); }
const int P = 1e9 + 7,N = 500050;
int n,fac[N],inv[N];
int x[N];
int f[N];
long long ans,now;
int main(){
	int i;
	read(n);
	for (i = 1; i <= n; ++i) read(x[i]);
	for (fac[0] = i = 1; i <= n; ++i) fac[i] = 1ll*fac[i-1]*i%P;
	for (inv[0] = inv[1] = 1,i = 2; i <= n; ++i) inv[i] = 1ll*inv[P%i]*(P-P/i)%P;
	for (i = 1; i <= n; ++i){
		int prob; prob = inv[i];
		f[i] = 1ll*prob*(f[i-1]+1) % P;
		f[i] += 1ll*(P+1-prob)*f[i-1] % P;
		f[i] %= P;
	}
	ans = 0;
	for (i = 1; i <= n-1; ++i){
		int dist = x[i+1] - x[i];
		ans = (ans + 1ll*dist*f[i]%P)%P;
	} 
	cout << 1ll*fac[n-1]*ans%P << endl;
    return 0;

(C) (Cookie) (Distribution)

题目链接

$description : $

有 (n) 个人 (,) 在 (k) 天中你要给他们发糖果 (.)

读入 (a_1 .. a_k,) 其中 (a_i) 表示第(i)天会从(n)个人当中均匀随机选出 (a_i) 个人 (,) 并给他们每人发一颗糖 (.)

记(c_i)为第(i)个人发到的糖果个数(,)求(Pi c_i)的期望(.)

(n<=10^3,k<=20)

(solution:)

求 (prod c_i) 相当于求 从 (n) 个人中每个人手里选一颗糖的方案数

考虑枚举 (x_i) 表示从第 (i) 个人手里选的糖果是从哪一天来的 (.)

(x_i) 是什么并不重要(,)重要的是(cnt2_i :) 表示有多少(x_j) (=) (i)

这种选法对答案的贡献是

[large prod^{k}_{i=1} C^{a_k-cnt2_k}_{n-cnt2_k} imes prod_{i=1}^{k} (cnt2_i!)^{-1} imes n! ]

然后用一个(O(nk^2) dp)就可以解决这个问题了(.)

(code :)

#include <bits/stdc++.h>
using namespace std;
inline int read(){
    int x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0',c = getchar();
    return x;
}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
inline void write(int x){if (x > 9) write(x/10); putchar(x%10+'0'); }

const int N = 1050,K = 20 + 5,P = 1e9 + 7;
int n,k,a[K];
int fac[N],nfac[N],inv[N];
inline int C(int n,int m){ return (n<m||n<0||m<0) ? 0 : 1ll*fac[n]*nfac[m]%P*nfac[n-m]%P; }
int t[K][N];
int dp[K][N];
int main(){
	int i,j,e;
	read(n),read(k); for (i = 1; i <= k; ++i) read(a[i]);
	inv[0] = fac[0] = nfac[0] = inv[1] = fac[1] = nfac[1] = 1;
	for (i = 2; i <= n; ++i){
		fac[i] = 1ll*fac[i-1]*i%P;
		inv[i] = 1ll*(P-P/i)*inv[P%i]%P;
		nfac[i] = 1ll*nfac[i-1]*inv[i]%P;
	} 
	for (i = 1; i <= k; ++i)
	for (j = 0; j <= n; ++j) t[i][j] = 1ll*C(n-j,a[i]-j)*nfac[j]%P;
	dp[0][0] = 1;
	for (i = 1; i <= k; ++i)
	for (j = 0; j <= n; ++j){
		dp[i][j] = 0;
		for (e = 0; e <= j; ++e) dp[i][j] = (dp[i][j] + 1ll * t[i][e] * dp[i-1][j-e]) % P;
	}
	int ans = 1ll * dp[k][n] * fac[n] % P; 
    cout << ans << endl;
	return 0;
}

(D) (Arrangement)

题目链接

(description :)

给你一张图(G,) (G)有(n)个点(,) (n*(n-2)) 条有向边(.)

图(G)的补图是一个所有点出度(=1) (() 可能有自环 ()) 的有向图(.)

求出一条字典序最小的哈密尔顿路径(,)如果不存在则输出(-1.)

(n<=10^5)

(solution :)

首先如果规模足够小((n<=8))就可以直接暴力(O(n!))

然后我们考虑对于(n)更大的情况来处理答案(.)

记$cnt_i = $ 当前的图中(,) 有多少(a_j = i)

如果存在一个点(p)满足(cnt_p = n-1)那么我们必须选这个点作为路径的第一个点(.)

如果不存在这样的一个点(,)我们就可以选择当前的点当中编号最小的点(.)

然后问题就变成了一个规模为(n-1)的子问题(，)只是多了一个开头不能为某个数的限制(.)

那么我们就可以写一个支持 查询(cnt)最大值(,) (cnt)单点修改 和 查询当前点集的编号最小者(,) 删除一个点的数据结构(,) 再写一个(O(n!))的暴力即可(.)

复杂度(O(8! +nlogn))

(code:)

#include <bits/stdc++.h>
using namespace std;
inline int read(){
    int x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0',c = getchar();
    return x;
}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
inline void write(int x){if (x > 9) write(x/10); putchar(x%10+'0'); }
const int N = 100050;
int n,a[N];

namespace subtask0{
	int ans[100],vis[100]; bool ok;
	int p[100];
	inline void chk(){
		for (int i = 2; i <= n; ++i) if (a[ans[i-1]] == ans[i]) return;
		ok = 1; for (int i = 1; i <= n; ++i) p[i] = ans[i];
		return; 
	}
	inline void dfs(int dep){
		if (ok) return;
		if (dep > n){ chk(); return; }
		for (int i = 1; i <= n; ++i) if (!vis[i]){
			vis[i] = 1,ans[dep] = i;
			dfs(dep+1);
			vis[i] = 0;
		}
	}
	inline void solve(){
		ok = 0; memset(vis,0,sizeof(vis)); memset(ans,0,sizeof(ans)); dfs(1);
		if (!ok){ puts("-1"); return; }
		for (int i = 1; i <= n; ++i) cout << p[i] << ((i<n) ? (' '):('
'));
	}
}

int mx[N<<2],mxi[N<<2],data[N<<2],cnt[N];
inline void Build(int o,int l,int r){
	if (l^r){
		int mid = l+r>>1; Build(o<<1,l,mid); Build(o<<1|1,mid+1,r);
		mx[o] = max(mx[o<<1],mx[o<<1|1]); data[o] = min(data[o<<1],data[o<<1|1]);
		mxi[o] = mxi[ (mx[o<<1]>mx[o<<1|1]) ? (o<<1) : (o<<1|1) ];
		return;
	}
	mx[o] = cnt[l]; data[o] = (mx[o] >= 0) ? (l) : (n+1); mxi[o] = l;
}
inline int Ask(int o,int l,int r,int p){
	if (l==r) return mx[o];
	int mid = l+r>>1; return (p<=mid) ? Ask(o<<1,l,mid,p) : Ask(o<<1|1,mid+1,r,p);
} 
inline void Add(int o,int l,int r,int p,int v){
	if (l==r){ mx[o]=v; data[o] = (mx[o] >= 0) ? (l) : (n+1); return; }
	int mid = l+r>>1; if (p<=mid) Add(o<<1,l,mid,p,v); else Add(o<<1|1,mid+1,r,p,v);
	mx[o] = max(mx[o<<1],mx[o<<1|1]); data[o] = min(data[o<<1],data[o<<1|1]);
	mxi[o] = mxi[ (mx[o<<1]>mx[o<<1|1]) ? (o<<1) : (o<<1|1) ];
}
inline int Nowv(){ return data[1]; }
inline int Query(int pos){ if (pos==0) return -1; return Ask(1,1,n,pos); }
inline void Modify(int x,int v){ cnt[x] = v; Add(1,1,n,x,v); }


int ans[N];
int vis[N],nowc[N],lc;
int qwq;
inline void chk(){
	for (int i = qwq; i <= n; ++i) if (ans[i]==a[ans[i-1]]) return;
	for (int i = 1; i <= n; ++i) write(ans[i]),putchar((i<n)?(' '):('
'));
	exit(0);
}
inline void dfs(int dep){
	if (dep>n){ chk(); return; }
	for (int i = 1; i <= lc; ++i) if (!vis[i]){
		vis[i] = 1;
		ans[dep] = nowc[i];
		dfs(dep+1);
		vis[i] = 0;
	}
}
inline void solve_force(int l,int r){
	qwq = l;
	for (int i = 1; i <= n; ++i) if (Query(i) >= 0) nowc[++lc] = i,vis[lc] = 0;
	dfs(l);
}
inline bool unproperty(){
	for (int i = 1; i <= n; ++i) cnt[i] = 0;
	for (int i = 1; i <= n; ++i) ++cnt[ans[i]];
	for (int i = 1; i <= n; ++i) if (cnt[i] != 1) return 1;
	for (int i = 2; i <= n; ++i) if (ans[i] == a[ans[i-1]]) return 1;
	return 0;
}
int main(){
	int i,banp,ret,siz,p;
	int pp,val;
	read(n);
	for (i = 1; i <= n; ++i) read(a[i]);
	for (i = 1; i <= n; ++i) if (a[i]==i) a[i]=0;
	for (i = 1; i <= n; ++i) ++cnt[a[i]];
//	if (n==2){ puts("-1"); return 0; }
	if (n<=8){ subtask0::solve(); return 0; }
	
	Build(1,1,n);
	for (banp = 0,siz = n,i = 1; i <= n; ++i,--siz){
		if (siz <= 6){
			solve_force(i,n); 
		}
		banp = a[ans[i-1]];
		if (banp != 0 && (ret=Query(banp)) >= 0){
			Modify(banp,-1);
			if (mx[1] == siz-1){
				p = mxi[1];
				ans[i] = p;
				Modify(ans[i],-1);
			}
			else{
				p = data[1];
				ans[i] = p;
				Modify(ans[i],-1);
			}
			pp = a[ans[i]];
			if ((val=Query(pp))>=0){ --val; Modify(pp,val); }
			Modify(banp,ret);
		}
		else{
			if (mx[1] == siz-1){
				p = mxi[1];
				ans[i] = p;
				Modify(ans[i],-1);
			}
			else{
				p = data[1];
				ans[i] = p;
				Modify(ans[i],-1);
			}
			pp = a[ans[i]];
			if ((val=Query(pp))>=0){ --val; Modify(pp,val); }
		}
	}
	if (unproperty()){ puts("-1"); return 0; }
	for (i = 1; i <= n; ++i) write(ans[i]),putchar((i<n)?(' '):('
'));
}

(E) (Span) (Covering)

题目链接

(description:)

有(n)条长度为(l_i)的线段(.)

你要把这些线段放到一个长为(X)的坐标轴上 (() (n <=100) (X <= 500) ())

并且满足一个条件(:) 不能覆盖到外面(,)也不能有地方没有被任何一条线段覆盖到(.)

求方案数(.)

(solution:)

首先把(l_i)从大到小排序(.)

考虑一个(dp:)

$f[i][j][k] = $ 前(i)个区间组成了(j)个相邻的开区间(,)其总长度为(k)的方案数(.)

转移有三种(:)

(1.) 新开一个区间(;(j->j+1))

(2.) 把当前的某一个区间和我新加进去的区间合并为一个区间(;(j->j))

(3.) 把两个相邻的区间通过我新加进去的区间合并成一个区间(.(j->j-1))

其中第(1)种转移的长度是确定的(,)为(k + len)

第(2) (3)种转移的长度是需要枚举的(.)

看起来复杂度是(O(n^2X^2),)但是在枚举到(i)的时候(,)只有(k*l_i <= j)的状态是有用的(.)

所以复杂度为(O(???) 能过 ()) (() 好像是 (O(nX^2)?) ())

(code:)

#include <bits/stdc++.h>
using namespace std;
inline int read(){
    int x = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) x = x * 10 + c - '0',c = getchar();
    return x;
}
template <typename T> void read(T &x){
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
	while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}
inline void write(int x){if (x > 9) write(x/10); putchar(x%10+'0'); }
const int N = 100 + 5,M = 1050,P = 1e9 + 7;
int n,m;

int f[N][M],g[N][M];
inline void upd(int &x,int y){ x+=y; x>=P?x-=P:0; x<0?x+=P:0; }
inline void DP(int L){
	int i,j,k;
//	cout <<"DP " << endl;
//	for (i = 1; i <= n; ++i,cout << endl)
//	for (j = 1; j <= m; ++j) cout <<f[i][j] << ' ';
	
	for (i = 1; i <= n; ++i)
	for (j = 1; j <= m; ++j) g[i][j] = f[i][j],f[i][j] = 0;
	for (i = 1; i <= n; ++i)
	for (j = 1; j <= m; ++j) if (g[i][j]>0){
		upd(f[i+1][j+L],1ll*(i+1)*g[i][j]%P);
		upd(f[i][j],1ll*g[i][j]*(P+j-i*(L-1))%P);
		for (k = 1; k < L; ++k) upd(f[i][j+k],2ll*g[i][j]%P*i%P);
		for (k = 0; k < L-1; ++k) upd(f[i-1][j+k],1ll*g[i][j]*(L-k-1)%P*(i-1)%P);
	}
}
int a[N];
int main(){
	int i,j;
	int rm;
	read(n),read(m); for (i = 1; i <= n; ++i) read(a[i]),++a[i]; rm = m; m += 1;
	sort(a+1,a+n+1); reverse(a+1,a+n+1);
	memset(f,0,sizeof(f)); f[1][a[1]] = 1;
	for (i = 2; i <= n; ++i) DP(a[i]);
	
//	cout <<"DP " << endl;
//	for (i = 1; i <= n; ++i,cout << endl)
//	for (j = 1; j <= m; ++j) cout <<f[i][j] << ' ';
	int ans = 0;
	for (i = 1; i <= 1; ++i) ans += f[i][rm+i],ans %= P;
    cout << ans << endl;
	return 0;
}