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题目大意：给出N个数，要求你将N个数分成K个集合，使每个集合的（最大值 - 最小值）^ 2和达到最小

解题思路：先排个序，从小打大排

设dp[i][j]为前i个数分成j个集合最小平方和 得到转移方程dp[i][j] = dp[k][j - 1] + (val[i] - val[k + 1]) ^ 2 val[i]为第i个数的值 设l > k，且点l比点k优 则dp[k][j - 1] + (val[i] - val[k + 1]) ^ 2 >= dp[l][j - 1] + (val[i] - val[l + 1]) ^ 2 化简得val[i] >= (dp[l][j - 1] + val[l+ 1] ^ 2 - dp[k][j - 1] - val[k + 1] ^ 2 ) / (2 ＊ val[l + 1] - 2 ＊ val[k + 1]) 因为val随着i的增大而增大，所以得到斜率方程

#include 
   
    #include 
    
     #include 
     
      using namespace std;typedef long long LL;const int N = 10010;const int M = 5010;LL val[N];LL dp[M][N];int n, m;int que[N];void init() {    scanf("%d%d", &n, &m);    for (int i = 1; i <= n; i++)        scanf("%lld", &val[i]);    sort(val + 1, val + 1 + n);}LL getUp(int l, int k, int j) {    return dp[j - 1][l] + val[l + 1] * val[l + 1] - (dp[j - 1][k] + val[k + 1] * val[k + 1]);}LL getDown(int l, int k) {    return 2 * (val[l + 1] - val[k + 1]);}void getDp(int i, int k, int j) {    dp[j][i] = dp[j - 1][k] + (val[i] - val[k + 1]) * (val[i] - val[k + 1]);}int cas = 1;void solve() {    for (int i = 1; i <= n; i++)        dp[1][i] = (val[i] - val[1]) * (val[i] - val[1]);    int head, tail;    for (int i = 2; i <= m; i++) {        head = tail = 0;        que[tail++] = i - 1;        for (int j = i; j <= n; j++) {            while (head + 1 < tail && getUp(que[head + 1], que[head], i) <= getDown(que[head + 1], que[head]) * val[j]) head++;            getDp(j, que[head], i);            while (head + 1 < tail && getUp(j, que[tail - 1], i) * getDown(que[tail - 1], que[tail - 2]) <= getUp(que[tail - 1], que[tail - 2], i) * getDown(j, que[tail - 1])) tail--;            que[tail++] = j;        }    }    if (m >= n) dp[m][n] = 0;    printf("Case %d: %lld\n", cas++, dp[m][n]);}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}
     
    
   

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