Atcoder Dwango Programming Contest 6th 题解
\(A\) \(Falling\) \(Asleep\)
\(description:\)
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;
}
\(B\) \(Fusing\) \(Slimes\)
\(description :\)
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;
\(C\) \(Cookie\) \(Distribution\)
$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
\[\large \prod^{k}_{i=1} C^{a_k-cnt2_k}_{n-cnt2_k} \times \prod_{i=1}^{k} (cnt2_i!)^{-1} \times n!
\]
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;
}
\(D\) \(Arrangement\)
\(description :\)
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) ? (' '):('\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)?(' '):('\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)?(' '):('\n'));
}
\(E\) \(Span\) \(Covering\)
\(description:\)
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;
}
