Codeforces Round #317 [AimFund Thanks-Round] (Div. 2) Minimization dp
2015-08-23 23:20
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原题链接:http://codeforces.com/contest/572/problem/D
题意
给你个数组A和n,k,问你排列A后,下面的最小值是多少。题解
先排个序,要填充像1,1+k,1+2k,1+3k....这样的序列,或像2,2+k,2+2k.......这样的序列,这些序列应该取排序数组中连续的一段才能使得答案最小,现在考察这些序列的大小,发现其大小要么是n/k,要么是n/k+1,所以可以dp[i][j]表示前 i 条序列我取了 j 个n/k这样的序列。转移就很简单了,详见代码。代码
#include <bits/stdc++.h> using namespace std; long long dp[5010][5010]; int n, k, a, b, x, y, z[300010]; long long inf = 1e18; int main() { //freopen("in.in", "r", stdin); //freopen("out.out", "w", stdout); while (scanf("%d %d", &n, &k) != EOF) { for (int i = 1; i <= n; i++) scanf("%d", &z[i]); sort(z + 1, z + n + 1); x = n / k; y = n / k + 1; b = n - k * x; a = k - b; dp[1][0] = z[y] - z[1]; dp[1][1] = z[x] - z[1]; for (int i = 2; i <= a + b; i++) { for (int j = 0; j <= min(a, i); j++) { if (i == j) { dp[i][j] = dp[i - 1][j - 1] + (long long)z[(i - 1) * x + x] - z[(i - 1) * x + 1]; } else { int tmp = j * x + (i - 1 - j) * y; if (tmp + y <= n) dp[i][j] = dp[i - 1][j] + (long long)z[tmp + y] - z[tmp + 1]; else dp[i][j] = inf; if (j > 0) { tmp = (j - 1) * x + (i - j) * y; if (tmp + x <= n) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + (long long)z[tmp + x] - z[tmp + 1]); } } } } //for (int i = 1; i <= 3; i++) printf("--> %I64d\n", dp[i][i]); printf("%I64d\n", dp[a + b][a]); } return 0; }
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