2022 ICPC 西安
https://codeforces.com/gym/104077
C. Clone Ranran
签到题
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
LL a, b, c;
LL divup(LL a, LL b){
if(a % b == 0)
return a / b;
return a / b + 1;
}
void sol(){
scanf("%lld%lld%lld", &a, &b, &c);
int k = 0, s = 1;
LL ans = c * b, tmp;
do{
k++;
s <<= 1;
tmp = k * a + divup(c, s) * b;
ans = min(ans, tmp);
// printf("k=%2d s=%3lld tmp=%lld\n", k, s, tmp);
}while(s <= c);
printf("%lld\n", ans);
// printf("\n");
}
int main(){
// freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
int T;
scanf("%d", &T);
while(T--)
sol();
return 0;
}
F. Hotel
签到题
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define ll long long
void solve();
int main(){
int T;T=1;
while(T--)solve();
return 0;
}
void solve(){
int n;
ll c1,c2,ans=0;
string s;
cin>>n>>c1>>c2;
for(int i=1;i<=n;i++){
cin>>s;
int sum=0;
if(s[0]==s[1]||s[0]==s[2]||s[1]==s[2])sum++;
if(sum)
ans+=c2+min(c2,c1);
else ans+=min(c1,c2)*3;
}if(c1*2<=c2){
cout<<n*3*c1;
return;
}
cout<<ans;
}
G. Perfect Word
分析：
很好想到对字符串按长度排序 (开始写错了就是因为按照字典序排序了)
依次考虑排序后的字符串
假如当前串为s[0,j]
如果s[0,j-1] 和 s[1,j] trie树中都出现过 那么我们将s[0,j]也加入trie树中
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define ll long long
const int maxn=1e5+5;
int tr[maxn][27],tot,cnt,ans=0;
string s[maxn],t;
void ins(string ch) {
int rt = 0;
int len=ch.size();
for (int i = 0; i < len; i++) {
if (!tr[rt][ch[i] - 'a'])
tr[rt][ch[i] - 'a'] = ++tot;
rt = tr[rt][ch[i] - 'a'];
}
}
bool find(string s)
{
int rt=0;
int len=s.size();
for(int i=0;i<len;i++)
{
if(!tr[rt][s[i]-'a']) return false;
rt=tr[rt][s[i]-'a'];
}
return true;
}
bool cmp(string aa,string bb){
if(aa.size()!=bb.size())return aa.size()<bb.size();
return aa<bb;
}
void solve();
int main(){
int T;T=1;
while(T--)solve();
return 0;
}
void solve(){
int n;
cin>>n;
for(int i=1;i<=n;i++){
cin>>t;
if(t.size()==1)ins(t),ans=1;
else s[++cnt]=t;
}
sort(s+1,s+1+cnt,cmp);
for(int i=1;i<=cnt;i++){
int j=s[i].size();
if(find(s[i].substr(1,j))&&find(s[i].substr(0,j-1)))
ans=max(ans,j),ins(s[i]);
}
cout<<ans;
}
L. tree
分析
不知道是我们太菜了还是太卷了 这道题我们是费尽脑汁才做出来的
开始想的是树形dp 发现方程都设不出来
换个想法 那就硬搞 首先想到把最长链给找出来 然后依次找除去最大链的次大
这个可以用长链剖分
对于这些长度降序的链 一定保证相同长度的链与链之间的点不存在祖先关系
如果不是先找最大再找次大而是随意遍历的话就可能存在祖先关系
假如总共有cnt条链 然后遍历一遍 前i个用满足第一个条件 需要用i个 后面所有的满足第二个条件 需要用 len[i+1]个
对于每种情况取个min即可
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define ll long long
const int N=1e6+5;
int n;
void solve();
void dfs(int,int);
bool cmp(int aa,int bb){
return aa>bb;
}
vector<int>Q[N];
int ans[N];
int cnt;
int main(){
int T;cin>>T;
while(T--)solve();
return 0;
}
int len[N], hson[N], top[N];
void dfs1(int p) {
len[p] = 1;
for (int i=0;i<Q[p].size();i++){
int q=Q[p][i];
dfs1(q);
if (len[q] + 1 > len[p])
hson[p] = q, len[p] = len[q] + 1;
}
}
void dfs2(int p, int tp) {
top[p] = tp;
if (hson[p]) dfs2(hson[p], tp);
for (int i=0;i<Q[p].size();i++){
int q=Q[p][i];
if (!top[q])
dfs2(q, q);
}
}
void cut() {
dfs1(1);
dfs2(1, 1);
}
void solve(){
scanf("%d",&n);cnt=0;
for(int i=1;i<=n;i++)Q[i].clear(),len[i]=hson[i]=top[i]=0;
for(int i=2,x;i<=n;i++)scanf("%d",&x),Q[x].push_back(i);
cut();
for(int i=1;i<=n;i++)
if(top[i]==i)ans[++cnt]=len[i];
sort(ans+1,ans+1+cnt,cmp);
int res=1e9+7;
ans[cnt+1]=0;
for(int i=0;i<=cnt;i++)
res=min(res,i+ans[i+1]);
cout<<res<<endl;
}
E. Find Maximum(待补)
分析
思路可能出现了问题 调了很久都没调出来
最后队友调处来了
#include<iostream>
using namespace std;
typedef long long ll;
ll f(ll x)
{
if (x == 0)
return 1;
else if (x % 3 == 0)
return f(x / 3) + 1;
else
return f(x - 1) + 1;
}
ll bacec = 0;//已有部分长度
ll bacec2 = 0;//最极端特例位置
int main()
{
int T;
cin >> T;
for (int I = 0; I < T; I++)
{
ll l, r;
cin >> l >> r;
//sizes = 0;
bacec = 0;//初始化多项式
bacec2 = 0;
ll m = 1;
//找多项式,bacec-1就是关键值
while (1)
{
//cout << bacec << " ";
if (bacec + m * 3 - 1 <= r)//需要扩大范围
{
m *= 3;
}
else
{
if (bacec + m * 2 - 1 < l)//在第二区内
{
//a[sizes] = m;
//t[sizes] = 2;
bacec += m * 2;
//sizes++;//计入项
m = 1;//m重置
continue;//继续找下一项
}
else if (bacec + m * 2 - 1 >= l && bacec + m * 2 - 1 <= r)//跨过第二标志
{
//a[sizes] = m;
//t[sizes] = 2;
bacec += m * 2;
//sizes++;//计入项
m = 1;//m重置
break;//结束
}
else if (bacec + m * 2 - 1 > r && bacec + m - 1 < l)//在第一区间
{
//a[sizes] = m;
//t[sizes] = 1;
bacec += m;
//sizes++;//计入项
m = 1;//m重置
continue;//继续找下一项
}
else if (bacec + m - 1 >= l && bacec + m - 1 <= r)//跨过第一标志//有特例
{
//a[sizes] = m;
//t[sizes] = 1;
ll temp = bacec + m * 2 - 1 - m/3;//最极端特例
bacec += m;
//sizes++;//计入项
m = 1;//m重置
if (temp >= l && temp <= r)//如果极端特例可取
{
bacec2 = temp;
}
break;//结束
}
}
}
cout << max(f(bacec - 1), f(bacec2)) << "\n";
}
}
B. Cells Coloring(待补)
分析：
很基础的网络流
#include <bits/stdc++.h>
using namespace std;
#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define fi first
#define se second
using ll = long long; using ull = unsigned long long; using pii = pair<int, int>;
template <typename A, typename B> bool smax(A &a, const B &b) { return a < b ? a = b, 1 : 0; }
template <typename A, typename B> bool smin(A &a, const B &b) { return b < a ? a = b, 1 : 0; }
constexpr int N = 250 * 2 + 2 + 7;
constexpr int M = 250 * 250 + 250 * 2 + 7;
constexpr int INF = INT_MAX;
int n, m, S, T, nod;
char s[N][N];
int sc[N], cl[N];
struct Edge {int to, ne, f;} g[M << 1]; int head[N], tot = 1;
void addedge(int x, int y, int z) { g[++tot].to = y; g[tot].f = z; g[tot].ne = head[x]; head[x] = tot; }
void adde(int x, int y, int z) { addedge(x, y, z); addedge(y, x, 0); }
int dis[N], gap[N], cur[N], q[N];
void bfs() {
int hd = 0, tl = 0;
q[++tl] = T, ++gap[dis[T] = 1];
while (hd < tl) {
int x = q[++hd];
for fec(i, x, y) if (!dis[y] && g[i^1].f) dis[y] = dis[x] + 1, ++gap[dis[y]], q[++tl] = y;
}
}
int dfs(int x, int a) {
if (x == T || !a) return a;
int flow = 0, f;
for (int &i = cur[x]; i; i = g[i].ne)
if (dis[x] == dis[g[i].to] + 1 && (f = dfs(g[i].to, std::min(a, g[i].f)))) {
g[i].f -= f, g[i ^ 1].f += f;
a -= f, flow += f;
if (!a) return flow;
}
--gap[dis[x]];
if (!gap[dis[x]]) dis[S] = nod + 1;
++gap[++dis[x]];
return flow;
}
int ISAP() {
static int ans = 0;
bfs();
while (dis[S] <= nod) memcpy(cur + 1, head + 1, sizeof(*head) * nod), ans += dfs(S, INF);
return ans;
}
int main() {
int c, d, mxk = 0, sum = 0;
cin >> n >> m >> c >> d;
for (int i = 1; i <= n; ++i) {
cin >> (s[i] + 1);
for (int j = 1; j <= m; ++j) s[i][j] = s[i][j] == '.', sc[i] += s[i][j], cl[j] += s[i][j], smax(mxk, cl[j]);
smax(mxk, sc[i]), sum += sc[i];
}
S = n + m + 1, T = nod = S + 1;
for (int i = 1; i <= n; ++i) adde(S, i, 0);
for (int i = 1; i <= m; ++i) adde(i + n, T, 0);
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j) if (s[i][j]) adde(i, j + n, 1);
ll ans = LLONG_MAX;
for (int k = 0; k <= mxk; ++k) {
if (k) {
memset(gap + 1, 0, sizeof(*gap) * nod);
memset(dis + 1, 0, sizeof(*dis) * nod);
for (int i = 1; i <= n; ++i) ++g[i * 2].f;
for (int i = 1; i <= m; ++i) ++g[(i + n) * 2].f;
}
smin(ans, (ll)c * k + (ll)d * (sum - ISAP()));
}
cout << ans << '\n';
return 0;
}
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