【AtCoder】ARC069
ARC069
C - Scc Puzzle
……不说了
#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);
}
int64 N,M;
void Solve() {
read(N);read(M);
if(N * 2 >= M) {out(M / 2);enter;}
else {out(N + (M - N * 2) / 4);enter;}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
D - Menagerie
容易发现固定两位后可以推出所有，枚举一下前两位是啥就行
#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[MAXN];
int num[MAXN];
void Solve() {
read(N);
scanf("%s",s);
for(int i = 0 ; i <= 1 ; ++i) {
for(int j = 0 ; j <= 1 ; ++j) {
num[0] = i;num[1] = j;
for(int k = 2 ; k < N ; ++k) {
num[k] = (s[k - 1] == 'x') ^ num[k - 2] ^ num[k - 1];
}
int t0 = num[N - 1] ^ num[1] ^ num[0] ^ (s[0] == 'x');
int tN = num[0] ^ num[N - 2] ^ num[N - 1] ^ (s[N - 1] == 'x');
if(t0 == 0 && tN == 0) {
for(int i = 0 ; i < N ; ++i) {
if(num[i] == 0) putchar('S');
else putchar('W');
}
enter;
return;
}
}
}
puts("-1");
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - Frequency
显然只有前缀最大值改变的位置会被统计
统计的方法是若从a变到b，我把b及以后所有的数大于a的全削成a是b的次数，统计起来可以直接把大于a的数全削成a，减去下一个位置大于b的数全削成b的次数
#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;
int64 a[MAXN],b[MAXN],pre[MAXN],ans[MAXN];
map<int,int64> zz,num;
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {read(a[i]);b[i] = a[i];ans[1] += a[i];}
sort(b + 1,b + N + 1);
for(int i = N ; i >= 1 ; --i) {
if(b[i] != b[i + 1]) {
num[b[i]] += N - i;
zz[b[i]] += zz[b[i + 1]];
}
num[b[i]]++;
zz[b[i]] += b[i];
}
int pos = 1;
for(int i = 1 ; i <= N ; ++i) {
if(a[i] > pre[i - 1] && i != 1) {
ans[i] += zz[pre[i - 1]] - pre[i - 1] * num[pre[i - 1]];
ans[pos] -= ans[i];
pos = i;
}
pre[i] = max(pre[i - 1],a[i]);
}
for(int i = 1 ; i <= N ; ++i) {
out(ans[i]);enter;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
F - Flags
大意是N个旗子可以放两个位置，设d是旗子之间最小的间距，d最大是多少
很套路。。。直接线段树优化建图跑2-SAT
#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);
}
struct tr_node {
int l,r,lc,rc;
}tr[MAXN * 10];
struct node {
int to,next;
}E[MAXN * 100];
int sumE,head[MAXN * 10];
int Ncnt,N;
pii val[MAXN * 2];
int p[MAXN][2],pos[MAXN],rt,x[MAXN][2],tot;
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void build(int &u,int l,int r) {
u = ++Ncnt;
tr[u].l = l;tr[u].r = r;
if(l == r) {pos[l] = u;return;}
int mid = (l + r) >> 1;
build(tr[u].lc,l,mid);
build(tr[u].rc,mid + 1,r);
add(u,tr[u].lc);add(u,tr[u].rc);
}
void Add_Range(int u,int l,int r,int v) {
if(l > r) return;
if(tr[u].l == l && tr[u].r == r) {add(v,u);return;}
int mid = (tr[u].l + tr[u].r) >> 1;
if(r <= mid) Add_Range(tr[u].lc,l,r,v);
else if(l > mid) Add_Range(tr[u].rc,l,r,v);
else {Add_Range(tr[u].lc,l,mid,v);Add_Range(tr[u].rc,mid + 1,r,v);}
}
int findL(int v) {
int l = 1,r = tot;
while(l < r) {
int mid = (l + r) >> 1;
if(x[val[mid].fi][val[mid].se] >= v) r = mid;
else l = mid + 1;
}
return l;
}
int findR(int v) {
int l = 1,r = tot;
while(l < r) {
int mid = (l + r + 1) >> 1;
if(x[val[mid].fi][val[mid].se] <= v) l = mid;
else r = mid - 1;
}
return l;
}
int dfn[MAXN * 10],low[MAXN * 10],sta[MAXN * 10],top,col[MAXN * 10],idx,instack[MAXN * 10],cor;
void Tarjan(int u) {
dfn[u] = low[u] = ++idx;
sta[++top] = u;
instack[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!dfn[v]) {
Tarjan(v);
low[u] = min(low[u],low[v]);
}
else if(instack[v] == 1) {
low[u] = min(low[u],dfn[v]);
}
}
if(low[u] == dfn[u]) {
++cor;
while(1) {
int x = sta[top--];
col[x] = cor;
instack[x] = 2;
if(x == u) break;
}
}
}
bool check(int d) {
Ncnt = 0,sumE = 0;
memset(head,0,sizeof(head));
build(rt,1,tot);
for(int i = 1 ; i <= tot ; ++i) {
int u = pos[i];
p[val[i].fi][val[i].se ^ 1] = u;
}
for(int i = 1 ; i <= tot ; ++i) {
int a,b;
a = findL(x[val[i].fi][val[i].se] - d + 1);
b = findR(x[val[i].fi][val[i].se] + d - 1);
Add_Range(1,a,i - 1,p[val[i].fi][val[i].se]);
Add_Range(1,i + 1,b,p[val[i].fi][val[i].se]);
}
idx = 0;top = 0;cor = 0;
memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));
memset(col,0,sizeof(col));memset(instack,0,sizeof(instack));
for(int i = 1 ; i <= N ; ++i) {
if(!dfn[i]) Tarjan(i);
}
for(int i = 1 ; i <= N ; ++i) {
if(col[p[i][0]] == col[p[i][1]]) return false;
}
return true;
}
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(x[i][0]);read(x[i][1]);
val[++tot] = mp(i,0);val[++tot] = mp(i,1);
}
sort(val + 1,val + tot + 1,[](pii a,pii b){return x[a.fi][a.se] < x[b.fi][b.se];});
int l = 0,r = 1000000000;
while(l < r) {
int mid = (l + r + 1) >> 1;
if(check(mid)) l = mid;
else r = mid - 1;
}
out(l);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
