题解:P9525 [JOISC2022] 团队竞技
涉及知识点:堆
解题思路
利用堆来维护每个能力的最大值。
如果当前的最大值都不属于同一只海狸,就输出。
否则把当重复的元素弹出堆。
如果堆空了还没有找到,说明没有,输出无解。
代码
#include <bits/stdc++.h>
#define int long long
#define ll __int128
#define db double
#define ldb long double
#define vo void
#define endl '\n'
#define il inline
#define re register
#define ve vector
#define p_q priority_queue
#define PII pair<int, int>
#define u_m unordered_map
#define bt bitset
using namespace std;
//#define O2 1
#ifdef O2
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#endif
struct IO {
#define MAXSIZE (1 << 20)
#define isdigit(x) (x >= '0' && x <= '9')
char buf[MAXSIZE], *p1, *p2;
char pbuf[MAXSIZE], *pp;
IO() : p1(buf), p2(buf), pp(pbuf) {}
~IO() {
fwrite(pbuf, 1, pp - pbuf, stdout);
}
char gc() {
if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin);
return p1 == p2 ? ' ' : *p1++;
}
bool blank(char ch) {
return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <class T>
void read(T &x) {
double tmp = 1;
bool sign = 0;
x = 0;
char ch = gc();
while (!isdigit(ch)) {
if (ch == '-') sign = 1;
ch = gc();
}
while (isdigit(ch)) {
x = x * 10 + (ch - '0');
ch = gc();
}
if (ch == '.') {
ch = gc();
while (isdigit(ch)) {
tmp /= 10.0, x += tmp * (ch - '0');
ch = gc();
}
}
if (sign) x = -x;
}
void read(char *s) {
char ch = gc();
for (; blank(ch); ch = gc());
for (; !blank(ch); ch = gc()) * s++ = ch;
*s = 0;
}
void read(char &c) {
for (c = gc(); blank(c); c = gc());
}
void push(const char &c) {
if (pp - pbuf == MAXSIZE) fwrite(pbuf, 1, MAXSIZE, stdout), pp = pbuf;
*pp++ = c;
}
template <class T>
void write(T x) {
if (x < 0) x = -x, push('-');
static T sta[35];
T top = 0;
do {
sta[top++] = x % 10, x /= 10;
} while (x);
while (top) push(sta[--top] + '0');
}
template <class T>
void write(T x, char lastChar) {
write(x), push(lastChar);
}
} io;
namespace COMB {
int fact[200000];
int Triangle[1010][1010];
void Fact(int n, int mod) {
fact[0] = 1;
for (int i = 1; i <= n; i ++ ) fact[i] = ((fact[i - 1]) % mod * (i % mod)) % mod;
}
void Pascal_s_triangle(int n, int mod) {
for (int i = 0; i <= n; i ++ ) Triangle[i][0] = 1;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= i; j ++ )
Triangle[i][j] = (Triangle[i - 1][j] + Triangle[i - 1][j - 1]) % mod;
}
int pw(int x, int y, int mod) {
int res = 1;
while (y) {
if (y & 1) res = ((res % mod) * (x % mod)) % mod;
x = (x % mod) * (x % mod) % mod;
y >>= 1;
}
return res;
}
int pw(int x, int y) {
int res = 1;
while (y) {
if (y & 1) res *= x;
x *= x;
y >>= 1;
}
return res;
}
int GCD(int x, int y, int mod) {
return __gcd(x, y) % mod;
}
int LCM(int x, int y, int mod) {
return (((x % mod) * (y % mod)) % mod / (GCD(x, y, mod) % mod)) % mod;
}
int C(int n, int m, int mod) {
if (m > n || m < 0) return 0;
return fact[n] * pw(fact[m], mod - 2, mod) % mod * pw(fact[n - m], mod - 2, mod) % mod;
}
int Ask_triangle(int x, int y) {
return Triangle[x][y];
}
}
using namespace COMB;
//#define fre 1
#define IOS 1
//#define multitest 1
const int N = 4e6 + 10;
const int M = 4e5 + 10;
const int inf = 1e12;
const int Mod = 1e9 + 9;
namespace zla {
int n;
struct node {
int a, b, c;
} A[N];
int vis[N];
p_q<PII> a, b, c;
il void Init() {
cin >> n;
for (int i = 1; i <= n; i ++ ) {
cin >> A[i].a >> A[i].b >> A[i].c;
a.push(make_pair(A[i].a, i));
b.push(make_pair(A[i].b, i));
c.push(make_pair(A[i].c, i));
}
}
il void Solve() {
for (int i = 1; i <= n; i ++ ) {
int x = a.top().second;
int y = b.top().second;
int z = c.top().second;
if (A[x].b == A[y].b || A[x].c == A[z].c) vis[x] = 1;
if (A[y].c == A[z].c || A[y].a == A[x].a) vis[y] = 1;
if (A[z].a == A[x].a || A[z].b == A[y].b) vis[z] = 1;
if (!vis[x] && !vis[y] && !vis[z]) {
cout << A[x].a + A[y].b + A[z].c;
return ;
}
while (!a.empty() && vis[a.top().second]) a.pop();
while (!b.empty() && vis[b.top().second]) b.pop();
while (!c.empty() && vis[c.top().second]) c.pop();
}
cout << -1;
}
il void main() {
Init();
Solve();
}
}
signed main() {
int T;
#ifdef IOS
ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
#ifdef fre
freopen("team.in", "r", stdin);
freopen("team.out", "w", stdout);
#endif
#ifdef multitest
cin >> T;
#else
T = 1;
#endif
while (T--) zla::main();
return 0;
}