BZOJ3928 [Cerc2014] Outer space invaders

第一眼，我勒个去。。。然后看到n ≤ 300的时候就2333了

首先把时间离散化，则对于一个时间的区间，可以知道中间最大的那个一定要被选出来，然后把区间分成左右两份

于是区间DP就好了，注意用左开右开的区间比较方便2333

如果暴力找区间内最大值是O(n3)的，当然st表就是O(n2logn)的了。。。不过st表辣么难蒟蒻根本不会QAQQQ

 

 1 /**************************************************************
 2     Problem: 3928
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:1820 ms
 7     Memory:2248 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <algorithm>
12  
13 using namespace std;
14 const int N = 605;
15 const int inf = 1e9;
16  
17 inline int read();
18  
19 struct data {
20     int st, ed, h;
21     data(int _s = 0, int _e = 0, int _h = 0) : st(_s), ed(_e), h(_h) {}
22      
23     inline void get() {
24         st = read(), ed = read(), h = read();
25     }
26 } a[N];
27  
28 int n, tot;
29 int tmp[N], len;
30 int f[N][N];
31  
32 int main() {
33     int T, H, i, j, k;
34     T = read();
35     while (T--) {
36         n = read();
37         for (i = 1; i <= n; ++i) {
38             a[i].get();
39             tmp[i * 2 - 1] = a[i].st, tmp[i * 2] = a[i].ed;
40         }
41         sort(tmp + 1, tmp + 2 * n + 1);
42         tot = unique(tmp + 1, tmp + 2 * n + 1) - tmp - 1;
43         for (i = 1; i <= n; ++i) {
44             a[i].st = lower_bound(tmp + 1, tmp + tot + 1, a[i].st) - tmp;
45             a[i].ed = lower_bound(tmp + 1, tmp + tot + 1, a[i].ed) - tmp;
46         }
47         tot += 1;
48         for (len = 0; len <= tot; ++len)
49             for (i = 0; i <= tot - len; ++i) {
50                 j = i + len, H = -1;
51                 for (k = 1; k <= n; ++k)
52                     if (i < a[k].st && a[k].ed < j && (H == -1 || a[H].h < a[k].h)) H = k;
53                 if (H == -1) f[i][j] = 0;
54                 else for (f[i][j] = inf, k = a[H].st; k <= a[H].ed; ++k)
55                     f[i][j] = min(f[i][j], a[H].h + f[i][k] + f[k][j]);
56             }
57         printf("%d\n", f[0][tot]);
58     }       
59     return 0;
60 }
61  
62 inline int read() {
63     static int x;
64     static char ch;
65     x = 0, ch = getchar();
66     while (ch < '0' || '9' < ch)
67         ch = getchar();
68     while ('0' <= ch && ch <= '9') {
69         x = x * 10 + ch - '0';
70         ch = getchar();
71     }
72     return x;
73 }
View Code

 

posted on 2015-04-19 16:28  Xs酱~  阅读(493)  评论(0)  编辑  收藏  举报