题意
给定 $n$ 个线段，每个线段有一个权值 $w_i$，每次取轴上一点，获得的权值为选择的覆盖在当前点上的线段的最大权值，求最终覆盖所有线段后需要的最小权值
题解
好吧是一道非常水的区间 $dp$ 然而我并没有想出来所以我是真的绝望
先把 $l_i, r_i$ 离散化后令 $f_{l, r}$ 表示完全覆盖区间 $[l, r]$ 的线段的最小权值，设完全处于 $[l, r]$ 的线段中权值取最大值的线段 $p$，则有转移方程
$$
f_{l, r} = \min \{f_{l, k - 1} + f_{k + 1, r} + w_p\} (l_p \le k \le r_p)
$$
所以注意一定要完全覆盖，$[l, k - 1], [k + 1, r]$ 才不会对 $[l, r]$ 造成影响，于是我就卡这儿了
谨以此文纪念我逝去的智商
代码
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| #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 600 + 10;
const int INF = 0x3f3f3f3f;
int f[MAXN][MAXN]= {0};
int T;
int N;
int l[MAXN], r[MAXN], value[MAXN];
int vcol[MAXN << 1], vcnt = 0;
int getnum () {
int num = 0;
char ch = getchar ();
while (! isdigit (ch))
ch = getchar ();
while (isdigit (ch))
num = (num << 3) + (num << 1) + ch - '0', ch = getchar ();
return num;
}
int main () {
T = getnum ();
for (int Case = 1; Case <= T; Case ++) {
N = getnum ();
vcnt = 0;
for (int i = 1; i <= N; i ++) {
l[i] = getnum (), r[i] = getnum (), value[i] = getnum ();
vcol[++ vcnt] = l[i], vcol[++ vcnt] = r[i];
}
sort (vcol + 1, vcol + vcnt + 1);
vcnt = unique (vcol + 1, vcol + vcnt + 1) - vcol - 1;
for (int i = 1; i <= N; i ++) {
l[i] = lower_bound (vcol + 1, vcol + vcnt + 1, l[i]) - vcol;
r[i] = lower_bound (vcol + 1, vcol + vcnt + 1, r[i]) - vcol;
}
for (int len = 2; len <= vcnt; len ++)
for (int i = 1; i <= vcnt - len + 1; i ++) {
int j = i + len - 1;
int maxv = - 1, sl, sr;
for (int k = 1; k <= N; k ++)
if (i <= l[k] && r[k] <= j)
if (value[k] > maxv) {
maxv = value[k];
sl = l[k], sr = r[k];
}
f[i][j] = maxv == - 1 ? 0 : INF;
if (maxv != - 1)
for (int k = sl; k <= sr; k ++)
f[i][j] = min (f[i][j], f[i][k - 1] + f[k + 1][j] + maxv);
}
printf ("%d\n", f[1][vcnt]);
}
return 0;
}
|
