Minimum Average Waiting Time

Problem Statement

Tieu owns a pizza restaurant and he manages it in his own way. While in a normal restaurant, a customer is served by following the first-come, first-served rule, Tieu simply minimizes the average waiting time of his customers. So he gets to decide who is served first, regardless of how sooner or later a person comes.

Different kinds of pizzas take different amounts of time to cook. Also, once he starts cooking a pizza, he cannot cook another pizza until the first pizza is completely cooked. Let's say we have three customers who come at time t=0, t=1, & t=2 respectively, and the time needed to cook their pizzas is 3, 9, & 6 respectively. If Tieu applies first-come, first-served rule, then the waiting time of three customers is 3, 11, & 16 respectively. The average waiting time in this case is (3 + 11 + 16) / 3 = 10. This is not an optimized solution. After serving the first customer at time t=3, Tieu can choose to serve the third customer. In that case, the waiting time will be 3, 7, & 17 respectively. Hence the average waiting time is (3 + 7 + 17) / 3 = 9.

Help Tieu achieve the minimum average waiting time. For the sake of simplicity, just find the integer part of the minimum average waiting time.

Input Format

• The first line contains an integer N, which is the number of customers.
• In the next N lines, the ith line contains two space separated numbers Ti and Li. Ti is the time when ith customer order a pizza, and Li is the time required to cook that pizza.

Output Format

• Display the integer part of the minimum average waiting time.

Constraints

• 1 ≤ N ≤ 105
• 0 ≤ Ti ≤ 109
• 1 ≤ Li ≤ 109

Note

• The waiting time is calculated as the difference between the time a customer orders pizza (the time at which they enter the shop) and the time she is served.

• Cook does not know about the future orders.

Sample Input #00

30 31 92 6

Sample Output #00

9

Sample Input #01

30 31 92 5

Sample Output #01

8

Explanation #01

Let's call the person ordering at time = 0 as A, time = 1 as B and time = 2 as C. By delivering pizza for A, C and B we get the minimum average wait time to be

(3 + 6 + 16)/3 = 25/3 = 8.33 

the integer part is 8 and hence the answer.

这道题我看了两个小时，十分简单的一道题，原因是我把题意给理解错了。我原先的理解是根据这怎样安排这N个顾客的服务顺序使得平均等待时间最短。其实问题不是这样的，问题是在当前时间t内，对所有已经到达的顾客中选择一个进行服务，最终使得平均等待时间最短，Heap的应用。

不能预知未来的程序：

#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <iostream>#include <algorithm>using namespace std;const int maxn = 100010;typedef long long ll;inline ll max(ll a, ll b) {    return a > b ? a : b;}struct Node1 {    ll a, b;    Node1() {}    Node1(ll t_a, ll t_b) : a(t_a), b(t_b) {}    bool operator < (const Node1 &p) const {        return a < p.a;    }}node[maxn];struct Node2 {    ll a, b;    Node2() {}    Node2(ll t_a, ll t_b) : a(t_a), b(t_b) { }    bool operator < (const Node2 &p) const {        return b > p.b;    }};int main() {    //freopen("aa.in", "r", stdin);    int n;    ll t = 0;    ll ans = 0;    int id = 1;    Node2 p;    priority_queue<Node2> pq;    scanf("%d", &n);    for(int i = 0; i < n; ++i) {        scanf("%lld %lld", &node[i].a, &node[i].b);        ans -= node[i].a;    }    sort(node, node + n);    t = node[0].a;    pq.push(Node2(node[0].a, node[0].b));    while(!pq.empty() || id < n) {        while(id < n) {            if(node[id].a <= t) {                pq.push(Node2(node[id].a, node[id].b));                id++;            } else {                break;            }        }        p = pq.top();        pq.pop();        t = t + p.b;        ans += t;    }    printf("%lld\n", ans / n);    return 0;}

能够预知未来的程序：

#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <iostream>#include <algorithm>using namespace std;const int maxn = 100010;typedef long long ll;struct Node1 {    ll a, b;    Node1() {}    Node1(ll t_a, ll t_b) : a(t_a), b(t_b) {}    bool operator < (const Node1 &p) const {        return a + b > p.a + p.b;    }};struct Node2 {    ll a, b;    Node2() {}    Node2(ll t_a, ll t_b) : a(t_a), b(t_b) {}    bool operator < (const Node2 &p) const {        return b > p.b;    }};inline ll max(ll a, ll b) {    return a > b ? a : b;}int main() {    priority_queue<Node1> pq1;    priority_queue<Node2> pq2;    int n;    ll a, b;    ll t = 0;    ll ans = 0;    Node1 p1;    Node2 p2;    scanf("%d", &n);    for(int i = 0; i < n; ++i) {        scanf("%lld %lld", &a, &b);        ans -= a;        pq1.push(Node1(a, b));    }    while(!pq1.empty() || !pq2.empty()) {        if(!pq1.empty()) {            if(pq1.top().a < t) {                while(!pq1.empty() && pq1.top().a < t) {                    p1 = pq1.top();                    pq1.pop();                    p2.a = p1.a;                    p2.b = p1.b;                    pq2.push(p2);                }                if(!pq1.empty()) {                    p1 = pq1.top();                    p2 = pq2.top();                    if(p1.a + p1.b <= t + p2.b) {                        if(p1.a + p1.b == t + p2.b && p1.b < p2.b) {                            t = t + p2.b;                            pq2.pop();                        } else {                            t = p1.a + p1.b;                            pq1.pop();                        }                    } else {                        t = t + p2.b;                        pq2.pop();                    }                } else {                    p2 = pq2.top();                    pq2.pop();                    t = t + p2.b;                }            } else {                p1 = pq1.top();                if(!pq2.empty()) p2 = pq2.top();                if(pq2.empty() || p1.a + p1.b <= t + p2.b) {                    if(!pq2.empty() && p1.a + p1.b == t + p2.b && p1.b < p2.b) {                        t = t + p2.b;                        pq2.pop();                    } else {                        pq1.pop();                        t = p1.a + p1.b;                    }                } else {                    t = t + p2.b;                    pq2.pop();                }            }            ans += t;            continue;        }        if(!pq2.empty()) {            p2 = pq2.top();            pq2.pop();            t = t + p2.b;            ans += t;        }    }    printf("%lld\n", ans / n);    return 0;}