[赛记] csp-s模拟3
奇观 55pts
赛时打的
考虑正解，首先，
对于
设
这东西可以前缀和维护，然后我们只需维护到
最后
点击查看代码
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define int long long
const long long mod = 998244353;
int n, m;
long long f[5][100005], g[5][100005];
vector<int> v[100005];
main() {
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
int x, y;
for (int i = 1; i <= m; i++) {
cin >> x >> y;
v[x].push_back(y);
v[y].push_back(x);
}
for (int i = 1; i <= n; i++) {
v[i].push_back(0);
v[i].push_back(i);
v[i].push_back(n + 1);
sort(v[i].begin(), v[i].end());
v[i].erase(unique(v[i].begin(), v[i].end()), v[i].end());
}
for (int i = 1; i <= n; i++) {
f[1][i] = 1;
g[1][i] = g[1][i - 1] + 1;
}
for (int i = 2; i <= 4; i++) {
for (int j = 1; j <= n; j++) {
for (int k = 1; k < v[j].size(); k++) {
f[i][j] += (g[i - 1][v[j][k] - 1] - g[i - 1][v[j][k - 1]] + mod) % mod;
f[i][j] = (f[i][j] + mod) % mod;
}
}
for (int j = 1; j <= n; j++) {
g[i][j] = g[i][j - 1] + f[i][j];
g[i][j] = (g[i][j] + mod) % mod;
}
}
long long ans = g[4][n] * g[4][n] % mod;
long long sum = 0;
for (int i = 1; i <= n; i++) {
long long su = v[i].size() - 2;
su = n - su;
sum = (sum + f[3][i] * su % mod * su % mod + mod) % mod;
}
ans = ans * sum % mod;
cout << ans;
return 0;
}
铁路 60pts
赛时说不清的乱搞；
其实并不需要真的加点，只需要用并查集维护一下哪些点是被删掉的，然后把这个连通块连到这个新点即可；
好吧，其实原题解写的就这么简洁，所以我也不想写了
点击查看代码
#include <iostream>
#include <cstdio>
using namespace std;
int n, m;
struct sss{
int t, ne;
}e[2000005];
int h[2000005], cnt;
void add(int u, int v) {
e[++cnt].t = v;
e[cnt].ne = h[u];
h[u] = cnt;
}
int f[500005][25];
int fa[1000005], dep[1000005];
int id[1000005];
int find(int x) {
if (x != fa[x]) fa[x] = find(fa[x]);
return fa[x];
}
void dfs(int x, int faa) {
f[x][0] = faa;
dep[x] = dep[faa] + 1;
for (int i = h[x]; i; i = e[i].ne) {
int u = e[i].t;
if (u == faa) continue;
dfs(u, x);
}
}
int ans;
int lca(int x, int y) {
if (x == y) return x;
if (dep[x] < dep[y]) swap(x, y);
for (int i = 20; i >= 0; i--) {
if (dep[f[x][i]] >= dep[y]) x = f[x][i];
}
if (x == y) return x;
for (int i = 20; i >= 0; i--) {
if (f[x][i] != f[y][i]) {
x = f[x][i];
y = f[y][i];
}
}
return f[x][0];
}
void w(int x, int y, int now) {
int sum = 0;
x = id[x];
y = id[y];
int lc = find(lca(x, y));
id[now] = lc;
while(find(x) != lc) {
fa[find(x)] = lc;
x = find(f[x][0]);
sum++;
}
while(find(y) != lc) {
fa[find(y)] = lc;
y = find(f[y][0]);
sum++;
}
ans -= sum;
}
int main() {
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
int x, y;
for (int i = 1; i <= n - 1; i++) {
cin >> x >> y;
add(x, y);
add(y, x);
}
for (int i = 1; i <= n + m; i++) {
fa[i] = i;
id[i] = i;
}
dfs(1, 0);
for (int j = 1; j <= 20; j++) {
for (int i = 1; i <= n; i++) {
f[i][j] = f[f[i][j - 1]][j - 1];
}
}
ans = n;
for (int i = 1; i <= m; i++) {
cin >> x >> y;
w(x, y, n + i);
cout << ans << '\n';
}
return 0;
}
光纤 10pts
赛时直接输出
考虑正解，需要用到一个东西叫旋转卡壳；
考虑最大的圆使其能够覆盖所有点，不难想到凸包（前提是你得学过），所以我们先求出凸包；
我们要考虑极限情况，即直线与圆相切的情况，那么我们最终的直线是要和至少一个圆相切的；
可以感性想一下，我们要求的这个直线是要把这个凸包“拦腰”截断的；
于是我们用旋转卡壳这个算法求出最长的点线距中的最小值，那么其一半的平方就是答案；
考虑证明一下，不难发现，把这条直线上下平移是更劣的，把其左右旋转会导致一边变大，一边变小；
最后，注意精度问题！，要用__int128，long double 会丢精度！
点击查看代码
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
#define int __int128
int n;
__int128 read() {
__int128 x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
void out(__int128 x) {
if (x < 0) putchar('-'), x = -x;
if (x > 9) out(x / 10);
putchar(x % 10 + '0');
}
struct poi{
int x, y;
}p[1000005], s[1000005];
int cnt;
struct vec{
int x, y;
inline int operator ^(const vec &A) const {
return x * A.y - y * A.x;
}
};
inline vec operator -(const poi x, const poi y) {
return vec{x.x - y.x, x.y - y.y};
}
struct line{
poi p;
vec v;
line(poi x, vec y) {
p = x;
v = y;
}
line(poi x, poi y) {
p = x;
v = y - x;
}
};
struct Q{
__int128 a, b;
long double val;
bool operator >=(const Q &A) const {
return val >= A.val;
}
};
inline int slen(vec x) {
return x.x * x.x + x.y * x.y;
}
inline bool cmp(poi x, poi y) {
vec a = x - p[1];
vec b = y - p[1];
if ((a ^ b) > 0.0) return 1;
else if ((a ^ b) == 0.0 && slen(a) <= slen(b)) return 1;
else return 0;
}
vector<poi> G() {
sort(p + 2, p + 1 + n, cmp);
s[++cnt] = p[1];
for (int i = 2; i <= n; i++) {
vec b = p[i] - s[cnt];
vec a = s[cnt] - s[cnt - 1];
while(cnt > 1 && (a ^ b) <= 0.0) {
cnt--;
b = p[i] - s[cnt];
a = s[cnt] - s[cnt - 1];
}
s[++cnt] = p[i];
}
s[++cnt] = p[1];
vector<poi> a;
a.clear();
for (int i = 1; i <= cnt; i++) a.push_back(s[i]);
return a;
}
__int128 aa, bb;
Q poi_to_line(poi x, line li) {
return Q{(__int128)((li.v ^ (x - li.p)) * (li.v ^ (x - li.p))), slen(li.v), ((li.v ^ (x - li.p)) * ((li.v ^ (x - li.p)))) / slen(li.v)};
}
void R(vector<poi> a) {
if (a.size() == 3) {
cout << "0/1";
exit(0);
}
int now = 0;
double sum = 999999999999999999999.00;
for (int i = 0; i < a.size() - 1; i++) {
line li(a[i], a[i + 1]);
while(poi_to_line(a[(now + 1) % a.size()], li) >= poi_to_line(a[now], li)) {
now = (now + 1) % a.size();
}
Q c = poi_to_line(a[now], li);
if (sum > c.val) {
sum = c.val;
aa = c.a;
bb = c.b;
}
}
}
signed main() {
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
n = read();
for (int i = 1; i <= n; i++) {
int x, y;
x = read();
y = read();
p[i].x = x;
p[i].y = y;
if (i != 1 && (p[i].y < p[1].y || (p[i].y == p[1].y && p[i].x < p[1].x))) swap(p[i], p[1]);
}
R(G());
bb *= 4;
__int128 gc = __gcd(aa, bb);
aa /= gc;
bb /= gc;
out(aa);
cout << '/';
out(bb);
return 0;
}
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