【AtCoder】Mujin Programming Challenge 2017
Mujin Programming Challenge 2017
A - Robot Racing
如果每個數都是一個一個間隔開的,那麽答案是\(n!\)
考慮把一個數挪到1,第二個數挪到3,以此類推,如果不行,證明前面中有個數肯定會被選擇,所以任意選一個數到終點,繼續這樣的操作
最後剩下的乘一個階乘即可
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 100005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } const int MOD = 1000000007; int N,ans; int x[MAXN]; int inc(int a,int b) { return a + b >= MOD ? a + b - MOD : a + b; } int mul(int a,int b) { return 1LL * a * b % MOD; } void update(int &x,int y) { x = inc(x,y); } void Solve() { read(N); for(int i = 1 ; i <= N ; ++i) read(x[i]); int pre = 0; int cnt = 0,ans = 1; for(int i = 1 ; i <= N ; ++i) { if(x[i] > pre) {++cnt;pre += 2;} else ans = mul(ans,cnt + 1); } for(int i = 1 ; i <= cnt ; ++i) ans = mul(ans,i); out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
B - Row to Column
顯然我們必須先恢復出一行來,然後去更新一開始沒有全滿的列
要麽是第j列有黑格子,我們用它去恢復出第j行,要麽給第j列創造一個黑格子,恢復出第j行
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 100005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int N; char s[505][505]; int sumr[505],sumc[505]; bool vis[505]; void Solve() { read(N); bool flag = 0; for(int i = 1 ; i <= N ; ++i) { scanf("%s",s[i] + 1); for(int j = 1 ; j <= N ; ++j) { if(s[i][j] == '#') flag = 1; } } if(!flag) { puts("-1");return; } for(int i = 1 ; i <= N ; ++i) { for(int j = 1 ; j <= N ; ++j) { if(s[i][j] == '#') vis[j] = 1; } } int cnt = 0; for(int j = 1 ; j <= N ; ++j) { int c = 0; for(int i = 1 ; i <= N ; ++i) { c += (s[i][j] == '#'); } if(c == N) ++cnt; } int ans = 2 * N; for(int i = 1 ; i <= N ; ++i) { int c = 0; for(int j = 1 ; j <= N ; ++j) { c += (s[i][j] == '.'); } ans = min(ans,c + N - cnt + (vis[i] ^ 1)); } out(ans);enter; } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
C - Robot and String
用處理出\(nxt(l)\)表示l之後到第幾個位置能變成空
同時記錄\(nxt_{a}(l),nxt_{b}(l)\)一直到z,初始時候先把s[l + 1]的數設成l + 1,然後從這個位置開始循環更新
然後我們一次詢問就是不斷的\(nxt(l - 1)\)嵌套,我們處理出2的次冪次操作即可
#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define mp make_pair #define pb push_back #define space putchar(' ') #define enter putchar('\n') #define eps 1e-10 #define MAXN 500005 //#define ivorysi using namespace std; typedef long long int64; typedef unsigned int u32; typedef double db; template<class T> void read(T &res) { res = 0;T f = 1;char c = getchar(); while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); } while(c >= '0' && c <= '9') { res = res * 10 +c - '0'; c = getchar(); } res *= f; } template<class T> void out(T x) { if(x < 0) {x = -x;putchar('-');} if(x >= 10) { out(x / 10); } putchar('0' + x % 10); } int N; char s[MAXN]; int nxt[MAXN][27],ri[MAXN][27]; void Solve() { scanf("%s",s + 1); N = strlen(s + 1); for(int i = 0 ; i <= N + 1; ++i) { for(int j = 0 ; j <= 26 ; ++j) { nxt[i][j] = N + 1; } } for(int i = 0 ; i <= N + 1 ; ++i) { for(int j = 0 ; j <= 19 ; ++j) ri[i][j] = N + 1; } for(int i = N - 1 ; i >= 0 ; --i) { int x = s[i + 1] - 'a'; nxt[i][x] = i + 1; for(int j = x + 1 ; j <= 26 ; ++j) { nxt[i][j] = nxt[nxt[i][j - 1]][j - 1]; } for(int j = 0 ; j < x ; ++j) { nxt[i][j] = nxt[nxt[i][26]][j]; } } for(int j = 0 ; j <= 19 ; ++j) { for(int i = N - 1 ; i >= 0 ; --i) { if(j == 0) ri[i][j] = nxt[i][26]; else ri[i][j] = ri[ri[i][j - 1]][j - 1]; } } int Q; read(Q); int l,r; for(int i = 1 ; i <= Q ; ++i) { read(l);read(r); --l; for(int j = 19 ; j >= 0 ; --j) { if(ri[l][j] <= r) l = ri[l][j]; } if(l == r) puts("Yes"); else puts("No"); } } int main() { #ifdef ivorysi freopen("f1.in","r",stdin); #endif Solve(); }
D - Oriented Tree
顯然D是直徑上取整
然後我們給每個點做一個標號,任取一個點\(h(v) = 0\)
如果\(u\rightarrow v\)有一條邊那麽\(h(v) - h(u) = 1\)
最後可以定義\(d(s,t) = (dist(s,t) + h(s) - h(t))/ 2\)
如果直徑是偶數,那麽可以發現
\(|h(s) - h(t)| \leq 2D - dist(s,t)\)
找到直徑的中心\(r\),距離中心距離為\(D\)的點,顯然標號應該一樣,假設標號都是0,我們可以得到任意一點u都有\(|h(u)| \leq D - dist(r,u)\)
這個是必要的,也是充分的,對於任意兩點\(u,v\)
\(|h(u) - h(v)| \leq |h(u)| + |h(v)| \leq 2D - dist(r,u) - dist(r,v) \leq 2D - dist(u,v)\)
所以我們認為\(|h(u)| \leq D - dist(r,u)\),做一個dp,可以得到所有解
如果直徑是奇數,就有了兩個中心,一個是s,一個是t
我們認為\(dist(u,s) < dist(t,u)\)的是白點,否則是黑點,那麽有兩種情況
如果u是白點,\(|h(u)| \leq D - 1 - dist(u,s)\)
如果u是黑點,\(|h(u)| \leq D - dist(u,t)\)
或者
u是白點,\(|h(u)| \leq D - dist(u,s)\)
u是黑點,\(|h(u)| \leq D - 1 - dist(u,t)\)
但是這兩種情況有交集
就是白點的距離s最遠的點全是0,另一邊距離最遠的全是-1,或者全是+1,這個時候我們如果所有點同時-1或者同時+1,會發現這兩種情況等價但是我們重復統計了
這個時候條件是
u是白點,\(|h(u)| \leq D - dist(u,s)\)
u是黑點,\(|h(u) + 1| \leq D - dist(u,t)\)
或者是
u是白點,\(|h(u)| \leq D - dist(u,s)\)
u是黑點,\(|h(u) - 1| \leq D - dist(u,t)\)
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 1005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007,V = 505;
struct node {
int to,next;
}E[MAXN * 2];
int sumE,head[MAXN];
int N;
int dis[MAXN],fa[MAXN],D;
int dp[MAXN][MAXN * 2];
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void update(int &x,int y) {
x = inc(x,y);
}
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void dfs(int u) {
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa[u]) {
fa[v] = u;
dis[v] = dis[u] + 1;
dfs(v);
}
}
}
void dfs1(int u,int lim,int on) {
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa[u]) {
dfs1(v,lim,on);
}
}
int l = dis[u] - lim - on,r = lim - dis[u] - on;
for(int i = l + V ; i <= r + V; ++i) {
dp[u][i] = 1;
for(int j = head[u] ; j ; j = E[j].next) {
int v = E[j].to;
if(v != fa[u]) {
int t = inc(dp[v][i - 1],dp[v][i + 1]);
dp[u][i] = mul(dp[u][i],t);
}
}
}
}
int Process_even(int rt) {
fa[rt] = 0;dis[rt] = 0;
dfs(rt);
memset(dp,0,sizeof(dp));
dfs1(rt,D,0);
int ans = 0;
for(int i = 0 ; i <= V * 2 ; ++i) {
update(ans,dp[rt][i]);
}
return ans;
}
int Process_odd(int s,int t,int on) {
fa[s] = t;fa[t] = s;dis[s] = 0;dis[t] = 0;
dfs(s);dfs(t);
memset(dp,0,sizeof(dp));
dfs1(s,D - 1,0);
dfs1(t,D,on);
int ans = 0;
for(int i = 1 ; i <= V * 2 ; ++i) {
update(ans,mul(dp[s][i],inc(dp[t][i - 1],dp[t][i + 1])));
}
return ans;
}
void Solve() {
read(N);
int a,b;
for(int i = 1 ; i < N ; ++i) {read(a);read(b);add(a,b);add(b,a);}
dis[1] = 0;fa[1] = 0;dfs(1);
int u = 1;
for(int i = 2 ; i <= N ; ++i) {
if(dis[i] > dis[u]) u = i;
}
dis[u] = 0;fa[u] = 0;
dfs(u);
u = 1;
for(int i = 2 ; i <= N ; ++i) {
if(dis[i] > dis[u]) u = i;
}
if(dis[u] % 2 == 0) {
int r = u;
for(int i = 1 ; i <= dis[u] / 2 ; ++i) r = fa[r];
D = dis[u] / 2;
int ans = Process_even(r);
out(ans);enter;
}
else {
int s = u;
for(int i = 1 ; i <= dis[u] / 2 ; ++i) s = fa[s];
int t = fa[s];
D = dis[u] / 2 + 1;
int ans = inc(Process_odd(s,t,0),Process_odd(t,s,0));
update(ans,MOD - Process_odd(s,t,-1));
update(ans,MOD - Process_odd(s,t,1));
out(ans);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
【AtCoder】Mujin Programming Challenge 2017