2018 计蒜之道初赛第一场

题解：典型的二分

 1 #pragma warning(disable:4996)
 2 #include<map>
 3 #include<queue>
 4 #include<string>
 5 #include<cstdio>
 6 #include<cstring>
 7 #include<iostream>
 8 #include<algorithm>
 9 #define ll long long 
10 using namespace std;
11 
12 const int maxn = 20005;
13 
14 ll s;
15 int n, p;
16 int a[maxn];
17 
18 bool check(int mid) {
19     int pos = lower_bound(a + 1, a + n + 1, mid) - a;
20     int ans = 0;
21     for (int i = pos; i <= n; i++) ans += (a[i] - mid);
22     if (ans > s / p) return false;
23     return true;
24 }
25 
26 int Solve() {
27     int l = 1, r = a[n];
28     for (int i = 1; i <= 30; i++) {
29         int mid = (l + r) >> 1;
30         if (check(mid)) r = mid;
31         else l = mid + 1;
32     }
33     return l;
34 }
35 
36 int main()
37 {
38     while (scanf("%d", &n) != EOF) {
39         for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
40         sort(a + 1, a + n + 1);
41         scanf("%d%lld", &p, &s);
42         printf("%d
", Solve());
43     }
44     return 0;
45 }

B-百度科学家

题解：数据比较小的时候建图，跑DFS统计总的污染值即可。

感受：题意不明！！！！！书替换后不会对以前的结果造成影响。所以~~要n何用。

 1 #pragma warning(disable:4996)
 2 #include<vector>
 3 #include<string>
 4 #include<cstdio>
 5 #include<cstring>
 6 #include<iostream>
 7 #include<algorithm>
 8 #define ll long long 
 9 using namespace std;
10 
11 const ll INF = 1000000000000000;
12 
13 int n, m;
14 int p[30], pos[30];
15 
16 vector<int> book[30];
17 
18 bool use[30];
19 
20 ll dp[30];
21 
22 void DFS(int u) {
23     use[u] = 1;
24     dp[u] = p[u];
25     for (int i = 0; i < book[u].size(); i++) {
26         int v = book[u][i];
27         if (!use[v]) {
28             DFS(v);
29             dp[u] += dp[v];
30         }
31     }
32 }
33 
34 int main()
35 {
36     while (cin >> n) {
37         for (int i = 1; i <= n; i++) pos[i] = i;
38         for (int i = 1; i <= n; i++) cin >> p[i];
39 
40         cin >> m;
41         int k = n;
42         for (int i = 1; i <= m; i++) {
43             int op;
44             cin >> op;
45             if (op == 0) {
46                 int x, v;
47                 cin >> x >> v;
48                 pos[x] = (++k);
49                 p[k] = v;
50             }
51             else {
52                 int x, l, r;
53                 cin >> x >> l >> r;
54                 for (int i = l; i <= r; i++) book[pos[x]].push_back(pos[i]);
55             }
56         }
57 
58         ll ans = INF;
59         for (int i = 1; i <= k; i++) {
60             memset(use, 0, sizeof(use));
61             memset(dp, 0, sizeof(dp));
62             DFS(i);
63             ans = min(ans, dp[i]);
64         }
65         cout << ans << endl;
66 
67         for (int i = 1; i <= k; i++) book[i].clear();
68     }
69     return 0;
70 }

2018 计蒜之道 初赛 第一场

2018 计蒜之道初赛第一场