HDU 6695 Welcome Party (greedy)

2019 Hang electric multi-school 101005

Topic links: HDU 6695

Game link: 2019 Multi-University Training Contest The 10

Problem Description

The annual welcome party of the Department of Computer Science and Technology is coming soon! Many students have been applying to show up at the welcome party, and every one of them can choose to sing a song or play crosstalk. This troubles the chief director a lot: how to arrange the program list, such that every student can have a chance to show up on the stage, and the satisfactory value of audiences is maximized?

To cope with this problem, the director proposes a model. In this model, every student has two attributes: the singing ability and crosstalking ability. The satisfactory value of audiences to singings is the maximum singing ability among all students that choose to sing a song; similarly, the satisfactory value to crosstalks is the maximum crosstalking ability among all students that choose play crosstalk. The strange thing is, the overall satisfactory value to the whole party is negatively related to the absolute difference between the satisfactory values to singings and crosstalks. The problem is, what is the minimum possible absolute difference between the satisfactory values of the two types of programs?

Note that:

every student should choose exactly one type of programs to play;

at least one student should sing a song, and at least one student should play crosstalk.

Input

The first line of input consists of a single integer \(T (1\le T\le 70)\), the number of test cases.

Each test case starts with a line of a single integer \(n (2\le n\le 100000)\), denoting the number of students applying to show up on the stage. Then follow \(n\) lines, each containing two integers \(x\) and \(y (0\le x,y\le 10^{18})\), denoting the singing ability and crosstalking ability of a student.

It is guaranteed that the sum of \(n\) over all test cases never exceeds \(1000000\).

Output

For each test case, output a single integer, denoting the minimum possible absolute difference between the satisfactory values of the two types of programs.

Sample Input

Sample Output

3
1

Solution

The meaning of problems

There \ (n \) individuals, \ (2 \) kinds of programs, each person should perform A program in which each program has at least one person show. With \ (x_i \) and \ (y_i \) represents \ (i \ (1 \ le i \ le n) \) the ability to value individual performances two kinds of programs. The ability now to make the show first show people the difference between the maximum and the maximum capacity of people performing the second program of the smallest, find the minimum value.

answer

greedy

Below, maintains two sets \ (S_1 \) and \ (S_2 \) .

Press \ (x \) descending enumeration. Suppose \ (x_i \) of \ (X \) maximum values (lower figure \ (4 \) ), the ratio of \ (x_i \) large choose \ (Y \) , i.e. take \ ( S_1 \) maximum values (lower figure \ (8 \) ). Ratio \ (x_i \) small ANDing \ (x_i \) closest \ (Y \) (in the figure below \ (3 \) ), because of the greater \ (Y \) can select \ (X \ ) (following figure \ (7 \) can be \ (2 \) replacement). Then take two of greater value to update \ (ANS \) (in the figure \ (| 3 - 4 | < | 8 - 4 | \) take \ (8--4 \) ), minimum maintenance \ (ANS \) can be.

Competition with teammate tree line before. After the game I \ (multiset \) wrote about.

title

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100000 + 5;
const ll inf = 2e18;

struct STU {
    ll x, y;
} s[maxn];
ll maxy[maxn];

int cmp(STU s1, STU s2) {
    return s1.x > s2.x;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while(T--) {
        int n;
        cin >> n;
        multiset<ll> s1, s2;
        for(int i = 0; i < n; ++i) {
            cin >> s[i].x >> s[i].y;
            s2.insert(s[i].y);
        }
        sort(s, s + n, cmp);
        ll ans = inf;
        for(int i = 0; i < n; ++i) {
            s2.erase(s2.find(s[i].y));  // 一个人只能选择一种表演
            if(!s1.empty()) {
                ans = min(ans, abs(*s1.rbegin() - s[i].x));
            }
            if(!s2.empty()) {
                multiset<ll>::iterator it = s2.lower_bound(s[i].x); // 找到第一个大于等于 s[i].x 的 y
                if(it == s2.end()) {
                    --it;
                }
                ll tmp = abs(*it - s[i].x);
                if(tmp < ans && (s1.empty() || *it > *s1.rbegin())) {
                    ans = tmp;
                }
                if(it != s2.begin()) {  // 找到最后一个小于 s[i].x 的 y
                    --it;
                    tmp = abs(*it - s[i].x);
                    if(tmp < ans && (s1.empty() || *it > *s1.rbegin())) {
                        ans = tmp;
                    }
                }
                s1.insert(s[i].y);
            }
        }
        cout << ans << endl;
    }
    return 0;
}