Codeforces Round #544 (Div. 3) 解題報告
摘要：
A. Middle of the Contest
#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define i...
A. Middle of the Contest
#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')
#define I_int ll
inline I_int read() {
I_int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
char F[200];
inline void write(I_int x) {
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0) {
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int
}
using namespace io;
using namespace std;
#define N 200100
int m1,m2,h1,h2;
int main() {
scanf("%d:%d", &h1, &m1);
scanf("%d:%d", &h2, &m2);
int s = h1 * 60 + m1, t = h2 * 60 + m2;
int mid = (s + t) >> 1, q1 = mid / 60, q2 = mid % 60;
if(q1 < 10) putchar('0'); printf("%d:",q1);
if(q2 < 10) putchar('0'); printf("%d",q2);
}
B. Preparation for International Women's Day
因為\(d_1+d_2=k\) ，所以\(d_1+d_2\%k=0\) 。
所以模一下存進對應剩餘系裡面，然後對每個對應的剩餘系統計一下即可。
#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')
#define I_int ll
inline I_int read() {
I_int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
char F[200];
inline void write(I_int x) {
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0) {
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int
}
using namespace io;
using namespace std;
#define N 200100
int a[N], n, k, vis[N];
int main() {
memset(vis, 0, sizeof(vis));
in(n); in(k);
for(int i = 1; i <= n; ++i) in(a[i]), a[i]%=k, vis[a[i]]++;
int ans = vis[0] / 2;
if(k % 2 == 0) ans += vis[k / 2] / 2;
for(int i = 1; i < (k + 1) / 2; ++i) {
int j = k - i;
ans += min(vis[i], vis[j]);
}
outn(ans*2);
}
C. Balanced Team
用雙指標搞搞就好了（因為有單調性）。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')
#define I_int ll
inline I_int read() {
I_int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
char F[200];
inline void write(I_int x) {
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0) {
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int
}
using namespace io;
using namespace std;
#define N 300010
int n, a[N];
int main() {
in(n);
for(int i = 1; i <= n; ++i) in(a[i]);
sort(a+1,a+n+1); int ans = 0;
for(int l = 1, r = 1; l <= n; ++l) {
while(a[r + 1] - a[l] <= 5 && r < n) ++r;
ans = max(ans, r - l + 1);
}
printf("%d\n", ans);
}
D. Zero Quantity Maximization
其實題意就是統計相同的分數有多少個（直接除gcd化成最簡分數），然後對於0 0要特殊統計（一定能計入答案）。
注意除0什麼的就行了，細節還是挺多的...
#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')
#define I_int ll
inline I_int read() {
I_int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
char F[200];
inline void write(I_int x) {
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0) {
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int
}
using namespace io;
using namespace std;
#define N 200100
int a[N], b[N], n, m;
map<pair<int, int>, int>mp;
int main() {
in(n); int ans = 0, s = 0;
for(int i = 1; i <= n; ++i) in(a[i]);
for(int i = 1; i <= n; ++i) {
int x = read(), g = __gcd(a[i], x);
if(!x && !a[i]) {
++s; continue;
}
if(!a[i]) continue;
ans = max(ans, ++mp[make_pair(a[i]/g, x/g)]);
}
printf("%d\n", ans+s);
}
E. K Balanced Teams
設\(f[i][j]\) 表示前i個人，選了j個team。
考慮繼承\(f[i][j]=f[i-1][j]\)
轉移則為\(f[i][j]=max\{f[k-1][j-1]+i-k+1\}(a[i]-a[k]<=5)\) ，
注意到k有單調性，那麼拿個指標維護一下最遠的k就可以優化到
\(O(n^2)\)
了。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')
#define I_int ll
inline I_int read() {
I_int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
char F[200];
inline void write(I_int x) {
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0) {
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int
}
using namespace io;
using namespace std;
#define N 5100
int vis[N], n, k;
ll a[N], f[N][N];
int main() {
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i) in(a[i]);
sort(a+1,a+n+1);
for(int j = 1; j <= k; ++j) {
for(int l = 1, i = 1; i <= n; ++i) {
f[i][j] = f[i - 1][j];
while(l < i && a[i] - a[l] > 5) ++l;
f[i][j] = max(f[i][j], f[l - 1][j - 1] + (i - l + 1));
}
}
printf("%lld\n", f[n][k]);
}
/*
f[i][j]位置i，選了j個team
*/
F1. Spanning Tree with Maximum Degree
題意殺型題目...題意是讓原來度數最大的點，弄出來生成樹後，度數還是和原來一樣，所以先把與度數最大的那個點連著的邊放進生成樹裡面，然後類似最小生成樹那樣用並查集判斷是否需要這條邊，選出n-1條即可。
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <map>
#include <set>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')
#define I_int ll
inline I_int read() {
I_int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
char F[200];
inline void write(I_int x) {
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0) {
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int
}
using namespace io;
using namespace std;
#define N 200100
int d[N], n, m;
struct edge {
int u, v;
}e[N];
vector<int>G[N];
int f[N], ans[N][2], cnt;
map<pair<int, int> , bool>mp;
int find(int x) {
if(f[x] == x) return x;
return f[x] = find(f[x]);
}
int main() {
in(n), in(m);
for(int i = 1; i <= m; ++i) {
in(e[i].u), in(e[i].v);
d[e[i].u]++; d[e[i].v]++;
G[e[i].u].push_back(e[i].v);
G[e[i].v].push_back(e[i].u);
if(e[i].u > e[i].v) swap(e[i].u, e[i].v);
}
int id = 0, sum = 0;
for(int i = 1; i <= n; ++i) {
f[i] = i;
if(d[i] > sum) sum = d[i], id = i;
}
for(int i = 0, len = G[id].size(); i < len; ++i) {
int x = find(id), y = find(G[id][i]);
f[y] = x;
x = id, y = G[id][i];
if(x > y) swap(x, y);
ans[++cnt][0] = x, ans[cnt][1] = y;
}
for(int i = 1; i <= m; ++i) {
if(cnt == n - 1) continue;
if(mp.count(make_pair(e[i].u, e[i].v))) continue;
int x = find(e[i].u), y = find(e[i].v);
if(x != y) {
f[y] = x;
ans[++cnt][0] = e[i].u;
ans[cnt][1] = e[i].v;
}
}
for(int i = 1; i <= cnt; ++i) printf("%d %d\n", ans[i][0], ans[i][1]);
return 0;
}
F2. Spanning Tree with One Fixed Degree
先把除了1以外的節點並起來，那麼就會出現幾個聯通塊，顯然必須保證1到每個聯通塊都至少有一條邊，那麼無解的判定條件就有了，必須有的邊不能超過d，1出去的邊也不能超過d。然後就按類似最小生成樹的流程把剩下的給並起來就好了
#include <bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define il inline
namespace io {
#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')
#define I_int ll
inline I_int read() {
I_int x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
x = x * 10 + c - '0';
c = getchar();
}
return x * f;
}
char F[200];
inline void write(I_int x) {
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0) {
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
}
#undef I_int
}
using namespace io;
using namespace std;
#define N 200100
int d[N], n, m;
struct edge {
int u, v;
}e[N];
int f[N], b[N][2], D, cnt, tot, ans[N][2], vis[N], t[N];
map<pair<int, int> , bool>mp;
int find(int x) {
if(f[x] == x) return x;
return f[x] = find(f[x]);
}
int main() {
in(n), in(m); in(D);
for(int i = 1; i <= n; ++i) f[i] = i;
for(int i = 1; i <= m; ++i) {
in(e[i].u), in(e[i].v);
if(e[i].u > e[i].v) swap(e[i].u, e[i].v);
if(e[i].u != 1 && e[i].v != 1) {
int x = find(e[i].u), y = find(e[i].v);
f[y] = x;
} else b[++tot][0] = e[i].u, b[tot][1] = e[i].v;
}
int now = 0; cnt = 0;
memset(vis, 0, sizeof(vis));
for(int i = 1; i <= tot; ++i) {
if(!vis[find(b[i][1])]) {
vis[find(b[i][1])] = 1;
ans[++now][0] = b[i][0], ans[now][1] = b[i][1];
mp[make_pair(b[i][0], b[i][1])] = 1;
}
}
if(tot < D) return puts("NO"), 0;
if(now > D) return puts("NO"), 0;
int l = 1;
for(int i = 1; i <= n; ++i) f[i] = i;
for(int i = 1; i <= now; ++i) {
f[find(ans[i][1])] = find(ans[i][0]);
}
for(int i = now + 1; i <= D; ++i, ++l) {
while(mp.count(make_pair(b[l][0], b[l][1])) && l < tot) ++l;
int x = find(b[l][0]), y = find(b[l][1]);
f[y] = x;
ans[i][0] = b[l][0], ans[i][1] = b[l][1];
}
now = D;
for(int i = 1; i <= m; ++i) {
if(now == n - 1) break;
if(e[i].u == 1 || e[i].v == 1) continue;
int x = find(e[i].u), y = find(e[i].v);
if(x != y) {
f[y] = x;
ans[++now][0] = e[i].u, ans[now][1] = e[i].v;
}
}
puts("YES");
for(int i = 1; i < n; ++i) printf("%d %d\n", ans[i][0], ans[i][1]);
return 0;
}