Sicily 1717. Computer

1717. Computer

Constraints

Time Limit: 2 secs, Memory Limit: 32 MB

Description

We often hear that computer is a magic, a great invention, or even a marvel. But actually computer is just a tool people use everyday. It is a machine that can help people to process many jobs effectively. Moreover, without computer, you can not play ICPC. So, guys, let’s study some stuff about computer here.
One computer has one CPU (Central Processing Unit). CPU can be idle or processing one job at any time. Jobs come randomly and are stored in the memory until finished. CPU will process jobs according to some strategies. The processing job can be interrupted and saved back so that CPU can be available for other jobs. 
Each job has a release time and a processing time. Assume that we know the schedule of all jobs, please generate a program to minimize the sum of completion times of all jobs using a strategy which assigns and interrupts jobs properly.
For example, suppose there are two jobs to be completed. Job 1 is released at time 1 and needs 4 time units to process. Job 2’s release time and processing time is 3 and 1. Figures below show three solutions:

Figure 1 shows a solution with the total complete time 4 + 6 = 10, and the result of Figure 2 and 3 are both 5 + 6 = 11. In fact, Figure 1 shows the optimal solution 
Please note that all of the jobs will be released, interrupted and assigned in integer time unit.

Input

Input may consist of multiple test cases.
Every test case begins with a line that contains one integer n (1<= n <= 50000) denoting the number of jobs. Each of the following n lines contains 2 integers: ri and pi, (1 <= ri <= 10^9, 1 <= pi <= 10000) denoting the release time and processing time of job i.
Input is terminated by EOF.

Output

For every test case, print one line with an integer denoting the minimum sum of completion times.

Sample Input

21 43 1

Sample Output

10

// Problem#: 1717// Submission#: 3585103// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <stdio.h>#include <algorithm>#include <queue>#include <vector>using namespace std;int n;struct node {    int start, len;};node arr[50100];bool cmp(const node & n1, const node & n2) {    if (n1.start != n2.start) return n1.start < n2.start;    return n1.len < n2.len;}void process() {    int i;    long long time, sum = 0;    for (i = 0; i < n; i++) scanf("%d%d", &arr[i].start, &arr[i].len);    sort(arr, arr + n, cmp);    time = arr[0].start;        priority_queue<int, vector<int>, greater<int> > q;    q.push(arr[0].len);        int index = 1;    while (1) {        while (!q.empty()) {            i = q.top();            q.pop();            if (index >= n) {                time += i;                sum += time;                continue;            }            if (time + i <= arr[index].start) {                time += i;                sum += time;                continue;            }            i = i - (arr[index].start - time);            q.push(i);            i = index;            time = arr[index].start;            while (1) {                if (arr[index].start != arr[i].start) break;                q.push(arr[index].len);                index++;                if (index >= n) break;            }        }        if (index < n) {            time = arr[index].start;            q.push(arr[index].len);            index++;        } else break;    }    printf("%lld\n", sum);}int main() {    while (scanf("%d", &n) != EOF) process();    return 0;}