2017 ACM/ICPC Asia Regional Qingdao Online
1001 Apple
给了四个点\(A,B,C,P\)，问点\(P\)是否在\(\Delta_{ABC}\)的外接圆外。
用C++被卡了精度，用Java过的。
import java.math.BigDecimal;
import java.util.Scanner;
/**
* Created by ToRapture on 2017/9/17.
*/
public class Main {
static BigDecimal dist(Point a, Point b) {
return a.x.subtract(b.x).multiply(a.x.subtract(b.x)).add(a.y.subtract(b.y).multiply(a.y.subtract(b.y)));
}
static BigDecimal circumscir(Point a, Point b, Point c, Point res) {
BigDecimal bx = b.x.subtract(a.x);
BigDecimal by = b.y.subtract(a.y);
BigDecimal cx = c.x.subtract(a.x);
BigDecimal cy = c.y.subtract(a.y);
BigDecimal d = BigDecimal.valueOf(2L).multiply(bx.multiply(cy).subtract(by.multiply(cx)));
BigDecimal x = cy.multiply( bx.multiply(bx).add(by.multiply(by))).subtract(by.multiply( cx.multiply(cx).add(cy.multiply(cy))))
.divide(d).add(a.x);
BigDecimal y = bx.multiply( cx.multiply(cx).add(cy.multiply(cy))).subtract(cx.multiply( bx.multiply(bx).add(by.multiply(by))))
.divide(d).add(a.y);
res.x = x;
res.y = y;
BigDecimal r = dist(a, res);
return r;
}
static int dcmp(BigDecimal x) {
BigDecimal eps = new BigDecimal("0.000000000000001");
if(x.compareTo(eps) > 0) return 1;
else if(x.compareTo(eps.multiply(new BigDecimal("-1"))) < 0) return -1;
else return 0;
}
static boolean fuck(Point a, Point b, Point c, Point p) {
if(p.equals(a) || p.equals(b) || p.equals(c)) return false;
Point o = new Point();
BigDecimal r = circumscir(a, b, c, o);
if(dcmp(dist(p, o).subtract(r)) <= 0) return false;
else return true;
}
static public void main(String[] args) {
Scanner scan = new Scanner(System.in);
int T = scan.nextInt();
while(--T >= 0) {
Point a = new Point();
Point b = new Point();
Point c = new Point();
Point p = new Point();
a.read(scan);
b.read(scan);
c.read(scan);
p.read(scan);
if(fuck(a, b, c, p)) System.out.println("Accepted");
else System.out.println("Rejected");
}
scan.close();
}
}
class Point {
BigDecimal x, y;
Point() {}
Point(BigDecimal x, BigDecimal y) {
this.x = x;
this.y = y;
}
public void read(Scanner scan) {
x = scan.nextBigDecimal();
y = scan.nextBigDecimal();
}
}
1003 The Dominator of Strings
Hash加乱搞过的，据说暴力strstr也能过。
#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long LL;
struct HASH {
typedef LL typec;
static const int MaxN = 100000 + 16;
static const LL key = 137;
typec H[MaxN], xp[MaxN];
void init(const char s[], int len) {
H[len] = 0;
for(int i = len - 1; i >= 0; --i) {
H[i] = H[i + 1] * key + s[i];
}
xp[0] = 1;
for(int i = 1; i <= len; ++i) {
xp[i] = xp[i - 1] * key;
}
}
typec get(int pos, int len) {
return H[pos] - H[pos + len] * xp[len];
}
} yoshiko;
const int N = 100000 + 16;
char buf[N];
LL H[N];
vector<int> F[256];
char fst[N];
int L[N];
int main(int argc, char **argv) {
int T; scanf("%d", &T);
while(T--) {
int n; scanf("%d", &n);
string str;
int select = -1;
for(int i = 0; i < n; ++i) {
scanf("%s", buf);
int l = strlen(buf);
fst[i] = buf[0];
L[i] = l;
if(str.size() < l) {
str = buf;
select = i;
}
LL h = 0;
for(int i = l - 1; i >= 0; --i) h = h * 137LL + buf[i];
H[i] = h;
}
yoshiko.init(str.c_str(), str.size());
for(char c = 'a'; c <= 'z'; ++c) F[c].clear();
for(int i = 0; i < str.size(); ++i) F[str[i]].push_back(i);
bool ok = true;
for(int i = 0; i < n && ok; ++i) {
if(i == select) continue;
ok = false;
for(auto p : F[fst[i]]) {
if(p + L[i] <= str.size() && yoshiko.get(p, L[i]) == H[i]) {
ok = true;
break;
}
}
}
if(ok) puts(str.c_str());
else puts("No");
}
return 0;
}
1008 Chinese Zodiac
签到题。
#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long LL;
int main(int argc, char **argv) {
map<string, int> mp;
mp["rat"] = 0;
mp["ox"] = 1;
mp["tiger"] = 2;
mp["rabbit"] = 3;
mp["dragon"] = 4;
mp["snake"] = 5;
mp["horse"] = 6;
mp["sheep"] = 7;
mp["monkey"] = 8;
mp["rooster"] = 9;
mp["dog"] = 10;
mp["pig"] = 11;
map<int, string> pm;
for(auto it : mp) pm[it.second] = it.first;
int T; cin >> T;
while(T--) {
string a, b;
cin >> a >> b;
int c = 0;
if(a == b) c = 12;
else {
int pos = mp[a];
while(pos != mp[b]) {
++c;
pos = pos + 1;
pos %= 12;
}
}
cout << c << endl;
}
return 0;
}
1009 Smallest Minimum Cut
求最小割中割边数最小的割的边数。
#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long LL;
struct Dinic { //时间复杂度O(min(n * sqrt(n), sqrt(m)) * m)
static const int maxn = 3000 + 16;
static const int INF = 0x7F7F7F7F;
struct Edge {
int from, to;
LL cap, flow;
Edge(int from, int to, LL cap, LL flow):
from(from), to(to), cap(cap), flow(flow) {}
Edge() {}
};
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn], cur[maxn];
int n, m, s, t; //n与m不需要特殊赋值，s与t在调用主过程的时候传就可以
void init(int n = maxn - 1) {
this->n = n;
for(int i = 0; i <= n; ++i) G[i].clear();
edges.clear();
}
void add_edge(int from, int to, LL cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
bool BFS() {
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while(!Q.empty()) {
int x = Q.front(); Q.pop();
for(int i = 0; i < G[x].size(); ++i) {
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow) {
vis[e.to] = true;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
LL DFS(int x, LL a) {
if(x == t || a == 0) return a;
LL flow = 0, f;
for(int &i = cur[x]; i < G[x].size(); ++i) {
Edge &e = edges[G[x][i]];
if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
LL MaxFlow(int s, int t) {
this->s = s; this->t = t;
LL flow = 0;
while(BFS()) {
memset(cur, 0, sizeof(cur));
flow += DFS(s, INF);
}
return flow;
}
} Dinic;
const LL BIG = 10000000LL;
int main(int argc, char **argv) {
int T; scanf("%d", &T);
while(T--) {
int n, m; scanf("%d%d", &n, &m);
int s, t; scanf("%d%d", &s, &t);
Dinic.init(n);
while(m--) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
Dinic.add_edge(u, v, c * BIG + 1);
}
printf("%lld\n", Dinic.MaxFlow(s, t) % BIG);
}
return 0;
}
1011 A Cubic number and A Cubic Number
好像也是签到题。
#include <bits/stdc++.h>
#define DBG(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long LL;
int main(int argc, char **argv) {
int T; cin >> T;
while(T--) {
double d; cin >> d;
double p = powf(d, 1.0 / 3);
LL s = p - 1;
if(s < 0) s = 0;
bool fuck = false;
for(LL i = s; ; i++) {
LL x = i * i * i;
LL y = (i + 1) * (i + 1) * (i + 1);
if(y - x == d) {
puts("YES");
fuck = true; break;
} else if(y - x > d) break;
}
if(!fuck) puts("NO");
}
return 0;
}
