https://www.bnuoj.com/bnuoj/problem_show.php?pid=51640
dp[i][j]表示前j个数,分成了i组,最小需要多少精力。
那么,求解订票dp[i][j]的时候,要么,第i组不做题,要么,第i组做1题、2题、3题....j题
先把数组排好序。然后暴力
dp[i][j] = dp[i - 1][j] //第i组不做题。
然后枚举一个k,表示[K.....j]是第i组做的题。那么就是dp[i - 1][k - 1] + (a[j] - a[k])^2
复杂度是O(n^3)
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 500 + 20;
int a[maxn];
LL dp[maxn][maxn];
void work() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
sort(a + 1, a + 1 + n);
for (int i = 1; i <= n; ++i) dp[0][i] = 1e18;
dp[0][0] = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
dp[i][j] = dp[i - 1][j];
for (int k = 1; k <= j; ++k) {
dp[i][j] = min(dp[i][j], dp[i - 1][k - 1] + 1LL * (a[j] - a[k]) * (a[j] - a[k]));
}
}
}
printf("%lld\n", dp[m][n]);
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
int t;
scanf("%d", &t);
while (t--) work();
return 0;
}
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