P1068 分數線劃定

結構體排序 自定義排序

 1 #include <cmath>
 2 #include <iostream>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 struct node {
 7     int id, score;
 8 }p[5005];
 9 
10 bool cmp(node x, node y) {
11     if (x.score == y.score) return x.id < y.id;
12     return x.score > y.score;
 
13 }
14 
15 int main() {
16     int n, m, idx;
17     cin >> n >> m;
18     idx = floor(m * 1.5); //向下取整
19     for (int i = 1; i <= n; i++) cin >> p[i].id >> p[i].score;
20     sort(p + 1, p + 1 + n, cmp);
21     cout << p[idx].score << " ";
22     int i = idx;
23
 
     while(p[i].score == p[i + 1].score && i <= n) i++;
24     cout << i << endl;
25     for (int j = 1; j <= i; j++) cout << p[j].id << " " << p[j].score << endl;
26     return 0;
27 }

P1803 凌亂的yyy / 線段覆蓋

結構體排序 貪心

 1 #include <cstdio>
 2
 
 #include <iostream>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 struct node {
 7     int a, b;
 8 }match[1000005];
 9 
10 bool cmp(node x, node y) {
11     return x.b < y.b;
12 }
13 
14 int main() {
15     int n, ans = 0, time = -1;
16     scanf("%d", &n);
17     for (int i = 0; i < n; i++) scanf("%d%d", &match[i].a, &match[i].b);
18     sort(match, match + n, cmp);
19     for (int i = 0; i < n; i++) {
20         if (match[i].a < time) continue;
21         ans++; time = match[i].b;
22     }
23     printf("%d\n", ans);
24     return 0;
25 }

P1208 [USACO1.3]混合牛奶 Mixing Milk

結構體排序 貪心

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int N = 2e6 + 10;
 6 
 7 struct milk {
 8     int p, a;
 9 };
10 
11 milk farmer[N];
12 
13 bool cmp(milk x, milk y) {
14     return x.p < y.p;
15 }
16 
17 int main() {
18     int n, m, ans = 0;
19     scanf("%d %d", &n, &m);
20     for (int i = 1; i <= m; i++) scanf("%d %d", &farmer[i].p, &farmer[i].a);
21     sort(farmer + 1, farmer + 1 + m, cmp);
22     for (int i = 1; i <= m; i++) {
23         if (farmer[i].a <= n) {
24             ans += farmer[i].p * farmer[i].a;
25             n -= farmer[i].a;
26         } else {
27             ans += farmer[i].p * n;
28             n = 0;
29         }
30         if (n <= 0) {
31             printf("%d\n", ans);
32             return 0;
33         }
34     }
35     printf("0\n");
36     return 0;
37 }