UVA1153-Keep the Customer Satisfied（贪心）

Posted 2021-01-08 npugen

Problem UVA1153-Keep the Customer Satisfied

Accept: 222  Submit: 1706
Time Limit: 3000 mSec

Problem Description

 

 

 Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
Data Each test case is described by one input ?le that contains all the relevant data: The ?rst line contains the number n of orders (n can be as large as 800000 for some test cases). It is followed by n lines. Each of which describes an order made of two integer values: the amount of steel (in tons) required for the order (lower than 1000) and its due date (in seconds; lower than 2×106).

 

 Output

For each test case, you are required to compute an optimal solution and your program has to write the number of orders that are accepted. The outputs of two consecutive cases will be separated by a blank line.

 

 Sample Input

1
6
7 15
8 20
6 8
4 9
3 21
5 22

 

Sample Output

4

 

题解：贪心算法，首先按截至时间从小到大排序，这个比较自然，然后贪心加区间，如果能直接加进去，就先加进去，如果不能，就比较该区间的持续时间和目前算进答案的持续时间最长的区间，如果该区间持续时间短，就删去大的，加进小的，正确性还是比较明显的。

 

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxn = 800000 + 100;
 6 
 7 struct Work {
 8     int q, d;
 9     Work() {}
10     Work(int _q, int _d) : q(_q), d(_d) {}
11     bool operator < (const Work &a)const {
12         if (a.d == d) return q < a.q;
13         else return d < a.d;
14     }
15 }work[maxn];
16 
17 int n;
18 
19 int main()
20 {
21     //freopen("input.txt", "r", stdin);
22     int iCase;
23     scanf("%d", &iCase);
24     bool flag = false;
25     while (iCase--) { 
26         scanf("%d", &n);
27         if (flag) printf("
");
28         flag = true;
29         int cnt = 0;
30         for (int i = 0; i < n; i++) {
31             scanf("%d %d", &work[i].q, &work[i].d);
32         }
33         
34         sort(work, work + n);
35         int ans = 0, pos = 0;
36         priority_queue<int> que;
37         for (int i = 0; i < n; i++) {
38             if (pos + work[i].q <= work[i].d) {
39                 pos += work[i].q;
40                 que.push(work[i].q);
41                 ans++;
42             }
43             else if(!que.empty()){
44                 int top = que.top();
45                 if (top > work[i].q) {
46                     pos += work[i].q - top;
47                     que.pop();
48                     que.push(work[i].q);
49                 }
50             }
51         }
52         printf("%d
", ans);
53     }
54     return 0;
55 }