1 #include<bits/stdc++.h>
 2 #define pb push_back
 3 typedef long long ll;
 4 using namespace std;
 5 typedef pair<int, int> pii;
 6 const int maxn = 1e3 + 10;
 7 int n;
 8 void solve() {
 9     map<int, int> ma;
10     scanf("%d", &n);
11     int x, y;
12     for (int i = 0; i < n; i++) {
13         scanf("%d%d", &x, &y);
14         if(ma.count(y)) {
15             ma[y] = max(ma[y], x);
16         } else ma[y] = x;
17     }
18     vector<pii> res;
19     auto it = ma.rbegin();
20     x = it->second - 1;
21     //cout << it->first <<endl;
22     while(it != ma.rend()) {
23         if(it->second > x) {
24             res.pb({it->second, it->first});
25             x = it->second;
26         }
27         it++;
28     }
29     for (pii t : res) {
30      printf("%d %d\n", t.first, t.second);
31     }
32 }
33
34 int main() {
35     freopen("test.in", "r", stdin);
36     //freopen("test.out", "w", stdout);
37     solve();
38     return 0;
39 }

View Code

#include<bits/stdc++.h>
#define pb push_back
typedef long long ll;
using namespace std;
typedef pair<int, int> pii;
const int maxn = 5e5 + 10;int n;
ll a[maxn];
ll s[maxn];
ll f(int x, int y) {return s[y] - s[x - 1];
}
ll res = 0;
void work(int x, int y) {if(x > y) return;if(x == y) {res = max(res, a[x] * a[x]);return;}int p = x;for (int i = x; i <= y; i++) {if(a[i] < a[p]) {p = i;}}res = max(res, a[p] * f(x, y));work(x, p - 1);work(p + 1, y);
}
void solve() {scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%lld", &a[i]);s[i] = s[i - 1] + a[i];}work(1, n);printf("%lld\n", res);
}int main() {freopen("test.in", "r", stdin);//freopen("test.out", "w", stdout);solve();return 0;
}

 1 #include<bits/stdc++.h>
 2 #define pb push_back
 3 typedef long long ll;
 4 using namespace std;
 5 typedef pair<int, int> pii;
 6 const int maxn = 3e3 + 10;
 7 struct node {
 8     int id, p, s, pr, t;
 9 };
10 int n, m, p;
11 node a[maxn];
12 //这里扩大范围
13 int f[maxn * 2];
14 int res[maxn];
15 class cmp {
16 public:
17     bool operator()(const node&x, const node&y) {
18         if(x.pr == y.pr) {
19             if(x.t == y.t) {
20                 return x.s < y.s;
21             } else {
22                 return x.t < y.t;
23             }
24         } else {
25             return x.pr > y.pr;
26         }
27     }
28 };
29 set<node, cmp> data[maxn];
30 void solve() {
31     int x, y, x1, y1;
32     scanf("%d%d%d", &n, &m, &p);
33     for (int i = 1; i <= p; i++) {
34         scanf("%d%d%d%d", &x, &y, &x1, &y1);
35         a[i] = {i, x, y, x1, y1};
36     }
37     sort(a + 1, a + p + 1, [](const node&x, const node&y)->bool{return x.s < y.s;});
38     int cur = 0;
39     f[1] = m;
40     set<int> se;
41     x = 1;
42     int cnt = 0;
43     //这里应该模拟到6000， 保证所有的任务都有机会执行，或者更大。
44     for (int i = 1; i <= 3000; i++) {
45         cur += f[i];
46         while(x <= p) {
47             //cout << x << endl;
48             if(a[x].s == i) {
49                 data[a[x].p ].insert(a[x]);
50                 //cout << "asd " << a[x].id << endl;
51                 x++;
52                 cnt++;
53             } else break;
54         }
55         if(cur <= 0 || cnt == 0) continue;
56         int t = cur;
57
58         for (int j = 0; j < t; j++) {
59             int id = -1;
60             int ti = 0;
61             for (int k = 1; k <= n; k++) {
62                 if(data[k].size() == 0)
63                     continue;
64                 else {
65                    // cout << k << " asd " << (data[k].begin())->t << endl;
66                     if(id == -1 || (data[k].begin())->t < ti) {
67                     id = k;
68                     ti = (data[k].begin())->t;
69                 }}
70             }
71             //cout << id << " " << ti << endl;
72             cur--;
73             cnt--;
74             node nd = *data[id].begin();
75             data[id].erase(data[id].begin());
76             res[nd.id] = i + nd.t;
77             f[i + nd.t]++;
78
79             if(cnt <= 0 || cur <= 0) break;
80         }
81
82     }
83     for (int i = 1; i <= p; i++)
84         printf("%d\n", res[i]);
85 }
86
87 int main() {
88     freopen("test.in", "r", stdin);
89     //freopen("test.out", "w", stdout);
90     solve();
91     return 0;
92 }