[BZOJ3244][NOI2013] 树的计数

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <iostream>
 4 #define MaxN 200010
 5 using namespace std;
 6 int n;
 7 int d[MaxN], b[MaxN], map[MaxN], s[MaxN], da[MaxN], fl[MaxN], pos[MaxN];
 8 
 9 void read(int &a){
10     a = 0; char c = getchar();
11     while (c < ‘0‘ || c > ‘9‘) c = getchar();
12     while (c >= ‘0‘ && c <= ‘9‘ ) a = a*10 + c-‘0‘, c = getchar();
13 }
14 
15 void Read_Data(){
16     read(n);
17     for (int i = 1; i <= n; i++) read(d[i]);
18     for (int i = 1; i <= n; i++) read(b[i]), map[b[i]] = i;
19     for (int i = 1; i <= n; i++) d[i] = map[d[i]];
20     for (int i = 1; i <= n; i++) pos[d[i]] = i;
21 }
22 
23 void Solve(){
24     da[1] = 1 , s[1] = 1, ++fl[1], --fl[2];
25     for (int i = 1; i < n; i++) {
26         if (pos[i] > pos[i+1]) da[i] = 1, ++fl[i], --fl[i+1];
27         s[i] = s[i-1]+da[i];
28     }
29     for (int i = 1; i < n; i++) {
30         if (d[i] > d[i+1]) continue;
31         int t = s[d[i+1]-1] - s[d[i]-1];
32         if (t) ++fl[d[i]], --fl[d[i+1]];
33     }
34     int flag = 0;
35     double ans = 0.0;
36     for (int i = 1; i < n; i++) {
37         flag += fl[i];
38         if (flag) ans += (double) da[i];
39         else ans += 0.5;
40     }
41     printf("%.3f\n", ans + 0.999);
42     printf("%.3f\n", ans + 1.000);
43     printf("%.3f\n", ans + 1.001);
44 }
45 
46 int main()
47 {
48     freopen("count.in", "r", stdin);
49     freopen("count.out", "w", stdout);
50     Read_Data();
51     Solve();
52     fclose(stdin);
53     fclose(stdout);
54     return 0;
55 }