【洛谷 P4072】 [SDOI2016]征途(斜率優化)
阿新 • • 發佈:2019-03-02
ret int 斜率 clas mem print play double memset
好久沒寫斜率優化板子都忘了,
硬是交了十幾遍。。
推一下柿子就能得到答案為
\[m*\sum x^2-(\sum x)^2\]
後面是個定值,前面簡單dp,斜率優化一下就行了。
\(f[i][j]=f[k][j-1]+sum[i]*sum[i]-2sum[i]sum[k]+sum[k]*sum[k]\)
\(-f[k][j-1]-sum[k]*sum[k]=-2sum[i]sum[k]-f[i][j]+sum[i]*sum[i]\)
#include <cstdio> #include <cstring> const int MAXN = 3010; int n, m; int f[MAXN][MAXN], sum[MAXN]; inline double k(int j, int i, int k){ return ((double)f[i][j - 1] + sum[i] * sum[i] - f[k][j - 1] - sum[k] * sum[k]) / ((double)sum[i] - sum[k]); } inline int min(int a, int b){ return a > b ? b : a; } int q[MAXN], head, tail; int main(){ scanf("%d%d", &n, &m); for(int i = 1; i <= n; ++i){ scanf("%d", &sum[i]); sum[i] += sum[i - 1]; } memset(f, 31, sizeof f); for(int i = 1; i <= n; ++i) f[i][1] = sum[i] * sum[i]; for(int j = 2; j <= m; ++j){ head = tail = 0; for(int i = 1; i <= n; ++i){ while(head < tail && k(j, q[head], q[head + 1]) < 2 * sum[i]) ++head; int K = q[head]; f[i][j] = f[K][j - 1] + (sum[i] - sum[K]) * (sum[i] - sum[K]); while(head < tail && k(j, q[tail - 1], q[tail]) >= k(j, q[tail], i)) --tail; q[++tail] = i; } } /*for(int j = 1; j <= m; ++j) for(int i = 1; i <= n; ++i) for(int k = 0; k < i; ++k) f[i][j] = min(f[i][j], f[k][j - 1] + (sum[i] - sum[k]) * (sum[i] - sum[k]));*/ printf("%d\n", m * f[n][m] - sum[n] * sum[n]); return 0; }
【洛谷 P4072】 [SDOI2016]征途(斜率優化)