【AtCoder】AGC029（A-E）
A - Irreversible operation
题解
把每个B后面的W个数累加起来即可
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
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);
}
char s[200005];
int64 ans = 0;
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
scanf("%s",s + 1);
int64 t = 0;
int N = strlen(s + 1);
for(int i = N ; i >= 1 ; --i) {
if(s[i] == 'W') ++t;
else ans += t;
}
out(ans);enter;
}
B - Powers of two
题解
这比赛打的太难受了
这个我花了一个树结构，蒙了一个dp，然后wa了，去硬上T2
最后发现我把一个N改成离散化后的tot就A了
就是考虑一个点，和它组成恰好大于它的\(2^{i}\)的值只有一个，构成了一片森林，对于每个森林从底向上更新，如果父亲和儿子能配对就尽可能多的配对
每个\(2^{i}\)的点若有剩余再两两配对
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
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 node {
int to,next;
}E[MAXN * 10];
map<int,int> mm;
int val[MAXN],A[MAXN],N,tot,ans;
int head[MAXN],sumE,dp[MAXN];
bool vis[MAXN];
int lowbit(int x) {
return x & (-x);
}
void add(int u,int v) {
E[++sumE].to = v;
E[sumE].next = head[u];
head[u] = sumE;
}
void Init() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
read(A[i]);
val[i] = A[i];
mm[A[i]]++;
}
sort(val + 1,val + N + 1);
tot = unique(val + 1,val + N + 1) - val - 1;
for(int i = 1 ; i <= N ; ++i) {
int a = val[i];
int k = lower_bound(val + 1,val + tot + 1,a) - val;
for(int64 t = 1 ; t <= 1073741824 ; t <<= 1) {
if(t > 2 * a) {
if(mm[t - a] == 0) continue;
int h = lower_bound(val + 1,val + tot + 1,t - a) - val;
add(k,h);
}
}
}
}
void dfs(int u) {
vis[u] = 1;
dp[u] = mm[val[u]];
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) {
dfs(v);
int t = min(dp[u],dp[v]);
dp[u] -= t;
dp[v] -= t;
ans += t;
}
}
}
void Solve() {
for(int i = 1 ; i <= tot ; ++i) {
if(!vis[i]) dfs(i);
if(lowbit(val[i]) == val[i]) ans += dp[i] / 2;
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}
C - Lexicographic constraints
题解
好不容易想到了一个分段维护一个字符个数进制下的数
没想到二分，gg了
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
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;
int A[MAXN];
int pw[MAXN][25],cnt[MAXN],ans;
vector<pii > v;
void Init() {
read(N);
for(int i = 1 ; i <= N ; ++i) read(A[i]);
cnt[1] = 1000000000;
for(int i = 1 ; i <= N ; ++i) {
pw[i][0] = 1;
for(int j = 1 ; j <= 20 ; ++j) {
if(pw[i][j - 1] > N / i) {
cnt[i] = j - 1;break;
}
pw[i][j] = pw[i][j - 1] * i;
}
}
}
bool check(int mid) {
v.clear();
v.pb(mp(A[1],0));int sum = A[1];
for(int i = 2 ; i <= N ; ++i) {
if(A[i] <= A[i - 1]) {
if(mid == 1) return false;
if(sum < A[i - 1]) {v.pb(mp(A[i - 1] - sum,0));sum = A[i - 1];}
int s = v.size() - 1;
for(int k = s ; k >= 0 ; --k) {
if(sum - v[k].fi >= A[i]) {sum -= v[k].fi;v.pop_back();}
else {
int t = sum - A[i];
if(t > cnt[mid]) v[k].se = 0;
else v[k].se /= pw[mid][t];
v[k].fi -= t;
sum = A[i];
}
if(sum == A[i]) break;
}
s = v.size() - 1;
for(int k = s ; k >= 0 ; --k) {
if(v[k].fi <= cnt[mid]) {
if(v[k].se + 1 < pw[mid][v[k].fi]) {++v[k].se;break;}
else {
v[k].se = 0;if(k == 0) return false;
}
}
else {++v[k].se;break;}
}
}
}
return true;
}
void Solve() {
int l = 1,r = N;
while(l < r) {
int mid = (l + r) >> 1;
if(check(mid)) r = mid;
else l = mid + 1;
}
out(l);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}
D - Grid game
题解
这题好水啊= =开D我就win了啊，sdC一直不过太难受了啊
就是考虑我们肯定是走到某个障碍物上面，游戏结束了
如何判断能不能走到呢，首先如果这一列是\(j\)，我们所在的行\(i\)必须大于等于\(j\)，如果有一个连成一串的障碍物从\((j,j)\)开始往下压
每次最多压了多少，就是当前列最靠下不能走的点\(h\),求一个\(t = h - i + 1\)
\(t\)每次取一个前缀\(max\)，然后对于每一列把行大于等于\(j + t\)的障碍物都设成不合法即可
把每一列能走到最靠上的障碍物取一个\(min\)
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
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 H,W,N;
vector<int> v[MAXN];
bool pass[MAXN];
void Solve() {
read(H);read(W);read(N);
int x,y;
for(int i = 1 ; i <= N ; ++i) {
read(x);read(y);
v[y].pb(x);
}
for(int i = 1 ; i <= W ; ++i) v[i].pb(H + 1);
sort(v[1].begin(),v[1].end());
int ans = v[1][0] - 1;
int t = 0;
for(int i = 2 ; i <= W ; ++i) {
sort(v[i].begin(),v[i].end());
int s = v[i].size();
int m = 0;
for(int k = 0 ; k < s ; ++k) {
if(v[i][k] <= i + t) {pass[v[i][k]] = 1;m = max(v[i][k],m);}
else if(pass[v[i][k] - 1]) {pass[v[i][k]] = 1;m = max(v[i][k],m);}
else {
ans = min(ans,v[i][k] - 1);
break;
}
}
t = max(t,m - i + 1);
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - Wandering TKHS
题解
关注一下一个路径的根缀最大值
把根缀最大值相同的部分挑出来
我们发现一个根缀最大值是\(u\)，\(u\)的某个儿子是\(v\)，\(v\)中的所有节点要走到\(u\)，就要经过\(v\)中所有根缀最大值为\(u\)的部分，这个可以差分实现
而根缀最大值自身讨论起来有点麻烦，直接搜索即可
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
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;
struct node {
int to,next;
}E[MAXN * 2];
int head[MAXN],sumE;
int mv[MAXN],siz[MAXN],d[MAXN],c[MAXN],son[MAXN];
void add(int u,int v) {
E[++sumE].next = head[u];
E[sumE].to = v;
head[u] = sumE;
}
void calc(int u,int fa,int val) {
son[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa && v < val) {
calc(v,u,val);
son[u] += son[v];
}
}
}
void dfs(int u,int fa) {
mv[u] = max(mv[fa],u);
siz[u] = 1;
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa) {
dfs(v,u);
siz[u] += siz[v];
d[u] += d[v];
}
}
if(mv[u] == u) {
calc(u,fa,mv[fa]);
d[u] = -siz[u];
}
if(mv[fa] == fa) d[u] += siz[u];
}
void dfs2(int u,int fa) {
if(fa) {
if(mv[u] == u) {c[u] += son[u];}
else if(mv[fa] == fa) {c[u] -= son[u];c[u] += d[u];}
c[u] += c[fa];
}
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa) {
dfs2(v,u);
}
}
}
void Init() {
read(N);
int x,y;
for(int i = 1 ; i < N ; ++i) {
read(x);read(y);add(x,y);add(y,x);
}
}
void Solve() {
dfs(1,0);
dfs2(1,0);
for(int i = 2 ; i <= N ; ++i) {
out(c[i]);
i == N ? enter : space;
}
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Init();
Solve();
}
F - Construction of a tree
\(2 \cdot 10^{5}\)的网络流，有毒，不会卡常，弃疗了，再见
