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 = 605;
const int inf = 1e9;

inline int read();

struct data {
int st, ed, h;
data(int _s = 0, int _e = 0, int _h = 0) : st(_s), ed(_e), h(_h) {}

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

int n, tot;
int tmp
, len;
int f

;

int main() {
int T, H, i, j, k;
T = read();
while (T--) {
n = read();
for (i = 1; i <= n; ++i) {
a[i].get();
tmp[i * 2 - 1] = a[i].st, tmp[i * 2] = a[i].ed;
}
sort(tmp + 1, tmp + 2 * n + 1);
tot = unique(tmp + 1, tmp + 2 * n + 1) - tmp - 1;
for (i = 1; i <= n; ++i) {
a[i].st = lower_bound(tmp + 1, tmp + tot + 1, a[i].st) - tmp;
a[i].ed = lower_bound(tmp + 1, tmp + tot + 1, a[i].ed) - tmp;
}
tot += 1;
for (len = 0; len <= tot; ++len)
for (i = 0; i <= tot - len; ++i) {
j = i + len, H = -1;
for (k = 1; k <= n; ++k)
if (i < a[k].st && a[k].ed < j && (H == -1 || a[H].h < a[k].h)) H = k;
if (H == -1) f[i][j] = 0;
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[0][tot]);
}
return 0;
}

inline int read() {
static int x;
static char ch;
x = 0, ch = getchar();
while (ch < '0' || '9' < ch)
ch = getchar();
while ('0' <= ch && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x;
}

View Code