codeforces747C Servers（思维）

There are n servers in a laboratory, each of them can perform tasks. Each server has a unique id — integer from 1 to n.

It is known that during the day q tasks will come, the i-th of them is characterized with three integers: ti — the moment in seconds in which the task will come, ki — the number of servers needed to perform it, and di — the time needed to perform this task in seconds. All ti are distinct.

To perform the i-th task you need ki servers which are unoccupied in the second ti. After the servers begin to perform the task, each of them will be busy over the next di seconds. Thus, they will be busy in seconds ti, ti + 1, ..., ti + di - 1. For performing the task, ki servers with the smallest ids will be chosen from all the unoccupied servers. If in the second ti there are not enough unoccupied servers, the task is ignored.

Write the program that determines which tasks will be performed and which will be ignored.

Input
The first line contains two positive integers n and q (1 ≤ n ≤ 100, 1 ≤ q ≤ 105) — the number of servers and the number of tasks.

Next q lines contains three integers each, the i-th line contains integers ti, ki and di (1 ≤ ti ≤ 106, 1 ≤ ki ≤ n, 1 ≤ di ≤ 1000) — the moment in seconds in which the i-th task will come, the number of servers needed to perform it, and the time needed to perform this task in seconds. The tasks are given in a chronological order and they will come in distinct seconds.

Output

Print q lines. If the i-th task will be performed by the servers, print in the i-th line the sum of servers' ids on which this task will be performed. Otherwise, print -1.

思路：

用一个数组vis记录每一台机器上次在哪次任务使用的，然后每次任务扫一遍多少个机器可用，如果小于这次任务的要求数， 就输出-1.

否则就从小到大找出这次任务需要的机器数，相加。

代码：

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <cctype>
#include <sstream>
#define INF 0x3f3f3f3f
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
typedef long long LL;
using namespace std;
const int maxn=100000+5;

struct Node
{
    int b,e,num;
}a[maxn];
int main()
{
    //freopen("in.txt", "r", stdin);
    int n,q,vis[maxn]={0};
    scanf ("%d%d",&n,&q);
    int t,k,d;
    for (int i=1;i<=q;i++) {
        scanf ("%d%d%d",&t,&k,&d);
        a[i].b=t;a[i].e=t+d;a[i].num=k;
        int x=0;
        for (int j=1;j<=n;j++){
            if (!vis[j] || a[vis[j]].e<=a[i].b) x++;//!vis[j]是指从未使用过的
        }
        if (x<a[i].num) printf ("-1\n");
        else {
            int ans=0;
            for (int j=1,m=0;j<=n&&m<a[i].num;j++){
                if (!vis[j] || a[vis[j]].e<=a[i].b) {
                    m++;
                    vis[j]=i;
                    ans+=j;
                }
            }
            printf ("%d\n",ans);
        }
    }
    return 0;
}