Flying Right POJ - 3038

Posted 2020-10-29 蒟蒻LQL

有一条从南到北的航线，航线上有N个机场1-n从南到北分布，每天早上飞机从1飞到n，傍晚从n飞到1。有k组乘客，他们数量为M[k]，从S飞到E，飞机上只有C个座位，计算每天飞机最多能拉多少乘客

贪心可以解决这个问题~（我一开始一直在想dp(lll￢ω￢)）

每个站点让所有乘客都上飞机，如果此时超载了，那么就让目的地离当前站点最远的乘客下飞机。可以用优先队列来维护。

 

emmm这个代码来自 https://blog.csdn.net/severus_qin/article/details/18956647，侵删

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <queue>
 6 #include <vector>
 7 using namespace std;
 8 const int maxn = 10010;
 9 struct event{
10     int t, c;
11     event(){}
12     event(int a, int b) : t(a), c(b){}
13     bool operator < (const event &rhs)const{
14         return t < rhs.t;
15     }
16 };
17 vector<event> v1[maxn], v2[maxn];
18 int k, n, c, num[2][maxn], ans;
19 int work(vector<event> vv[], int k){
20     priority_queue<event> que;
21     int res = 0, tmp = 0;
22     for (int i = 1; i <= n; i++){
23         res += num[k][i];
24         tmp -= num[k][i];
25         for (int j = 0; j < vv[i].size(); j++){
26             tmp += vv[i][j].c;
27             que.push(vv[i][j]);
28         }
29         while(tmp > c){
30             event tt = que.top(); que.pop();
31             if (tmp - c >= tt.c){
32                 tmp -= tt.c;
33                 num[k][tt.t] -= tt.c;
34             }else{
35                 num[k][tt.t] -= (tmp - c);
36                 tt.c -= (tmp - c);
37                 tmp = c;
38                 que.push(tt);
39             }
40         }
41     }
42     return res;
43 }
44 void solve(){
45     memset(num, 0, sizeof(num));
46     for (int i = 0; i < maxn; i++){
47         v1[i].clear(); v2[i].clear();
48     }
49     for (int i = 0; i < k; i++){
50         int x, y, z;
51         scanf("%d%d%d", &x, &y, &z);
52         if (x < y){
53             v1[x].push_back(event(y, z));
54             num[0][y] += z;
55         }else{
56             x = n - x + 1; y = n - y + 1;
57             v2[x].push_back(event(y, z));
58             num[1][y] += z;
59         }
60     }
61     ans = 0;
62     ans += work(v1, 0); ans += work(v2, 1);
63     printf("%d\n", ans);
64     return;
65 }
66 int main(){
67     while(scanf("%d%d%d", &k, &n, &c) == 3) solve();
68     return 0;
69 }

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