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

 /**************************************************************
     Problem: 3928
     User: rausen
     Language: C++
     Result: Accepted
     Time:1820 ms
     Memory:2248 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>

 using namespace std;
 const int N = ;
 const int inf = 1e9;

 inline int read();

 struct data {
     int st, ed, h;
     data(int _s = , int _e = , int _h = ) : st(_s), ed(_e), h(_h) {}

     inline void get() {
         st = read(), ed = read(), h = read();
     }
 } a[N];

 int n, tot;
 int tmp[N], len;
 int f[N][N];

 int main() {
     int T, H, i, j, k;
     T = read();
     while (T--) {
         n = read();
         for (i = ; i <= n; ++i) {
             a[i].get();
             tmp[i *  - ] = a[i].st, tmp[i * ] = a[i].ed;
         }
         sort(tmp + , tmp +  * n + );
         tot = unique(tmp + , tmp +  * n + ) - tmp - ;
         for (i = ; i <= n; ++i) {
             a[i].st = lower_bound(tmp + , tmp + tot + , a[i].st) - tmp;
             a[i].ed = lower_bound(tmp + , tmp + tot + , a[i].ed) - tmp;
         }
         tot += ;
         for (len = ; len <= tot; ++len)
             for (i = ; i <= tot - len; ++i) {
                 j = i + len, H = -;
                 for (k = ; k <= n; ++k)
                     if (i < a[k].st && a[k].ed < j && (H == - || a[H].h < a[k].h)) H = k;
                 if (H == -) f[i][j] = ;
                 else for (f[i][j] = inf, k = a[H].st; k <= a[H].ed; ++k)
                     f[i][j] = min(f[i][j], a[H].h + f[i][k] + f[k][j]);
             }
         printf("%d\n", f[][tot]);
     }
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }