[IOI2013]robots 题解
首先题意可以转化为\(:\)每个玩具有两个属性值\(x_i,y_i,\)弱机器人可以拿掉\(x_i<A_i\)的玩具\(,\)小机器人可以拿掉\(y_i<B_i\)的玩具\(.\)
不难想到可以二分答案\(.\)
考虑如何\(check\)是否能在\(mid\)分钟的时间内拿走所有的玩具\(.\)
首先如果只有一维\(,\)显然可以直接对\(x_i\)和\(A_i\)排序然后丢掉多余的机器人\(,\)直接比大小即可\(.\)
那么多加了一维就是对\(x\)先贪心\(,\)每次尽量拿走\(y\)更大的数字即可\(.\)
这个贪心很容易用一个堆来实现\(.\)
\(O(n\log ans \log n).\)
代码 \(:\)
#include <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &x){
x = 0; char ch = getchar();
while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - '0',ch = getchar();
}
const int N = 500005,M = 100005;
struct st{ int x,y; bool operator < (const st w) const{ return x < w.x; } }p[N];
int A,B,n,a[M],b[M],ca[M],cb[M]; priority_queue<int>H;
inline bool check(int Mid){
int i;
for (i = 1; i <= A; ++i) ca[i] = Mid;
for (i = 1; i <= B; ++i) cb[i] = Mid;
int now = 1;
while (!H.empty()) H.pop();
for (i = 1; i <= A; ++i){
while (now <= n && p[now].x < a[i]) H.push(p[now].y),++now;
while (ca[i]-- && !H.empty()) H.pop();
}
while (now <= n) H.push(p[now].y),++now;
for (i = B; i >= 1; --i)
while (!H.empty() && cb[i]-- && b[i] > H.top()) H.pop();
return H.empty();
}
int main(){
int i;
read(A),read(B),read(n);
for (i = 1; i <= A; ++i) read(a[i]); sort(a+1,a+A+1);
for (i = 1; i <= B; ++i) read(b[i]); sort(b+1,b+B+1);
for (i = 1; i <= n; ++i) read(p[i].x),read(p[i].y); sort(p+1,p+n+1);
int Ans = -1,L = 1,R = n,Mid;
while (L <= R){ Mid = L+R>>1; if (check(Mid)) R = Mid-1,Ans = Mid; else L = Mid + 1; }
cout << Ans << '\n';
return 0;
}