2020 ICPC南京 L - Let's Play Curling(离散化、模拟)
题目大意:
红队n个冰壶、蓝队m个冰壶,给出所有冰壶的坐标,找到一个位置c使得红队能赢且得分尽可能多,若红队能赢输出最多能得到的分数,若红队不能赢输出Impossible。
思路:
可将题意转化为求任意相邻的两个蓝队冰壶之间最多能有几个红队的冰壶。
通过upper_bound和lower_bound可以得到两个蓝队冰壶之间的左右边界。
注意边界的处理情况,如一组样例:
1
3
1
3 4 5
2
那么可将蓝队的冰壶看作是2和INF两个冰壶,求在这两个蓝队冰壶之间的红队冰壶数量,因此我们将蓝队的冰壶加入一个负无穷和正无穷来方便模拟。
Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL, LL> PI;
const double eps = 1e-6;
const LL N = 200010;
const LL INF = 0x3f3f3f3f;
const LL mod = 1000000007; //998244353
LL powmod(LL a, LL b) { LL res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res * a % mod; a = a * a % mod; }return res; }
LL a[N], b[N];
void init() {
for (LL i = 0; i < N; i++) {
a[i] = b[i] = 0;
}
}
int main() {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
LL T; cin >> T;
while (T--) {
init();
LL n, m; cin >> n >> m;
LL ans = 0;
for (LL i = 1; i <= n; i++) {
cin >> a[i];
}
for (LL i = 1; i <= m; i++) {
cin >> b[i];
}
b[0] = -INF;
b[m + 1] = INF;
sort(a + 1, a + 1 + n);
sort(b + 1, b + 1 + m);
for (LL i = 1; i <= m + 1; i++) {
LL lf = upper_bound(a + 1, a + 1 + n, b[i - 1]) - a; //两个blue石头之间
LL rt = lower_bound(a + 1, a + 1 + n, b[i]) - a - 1;
ans = max(ans, rt - lf + 1);
}
if (ans <= 0) cout << "Impossible" << endl;
else cout << ans << endl;
}
return 0;
}