HDU 3507 Print Article 斜率优化DP

来源：互联网发布：淘宝内衣模特方祺媛编辑：程序博客网时间：2024/05/21 09:50

题目大意：给出一个序列，可以连续出一段序列[l,r]，费用为sum[j][i] ^2+M，求输出整个序列的最小费用。

思路：裸DP方程：f[i] = f[j] + (sum[i] - sum[j - 1]) ^ 2 + M，然后整理一下斜率优化

=> f[j] + sum[j]^2 = 2 * sum[i] * sum[j] - M - f[i]

y = f[j] + sum[j] ^ 2

k = 2 * sum[i]

x = sum[j]

CODE：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 500010#define INF 1e15using namespace std;struct Point{long long x,y;Point(long long _ = 0,long long __ = 0):x(_),y(__) {}}q[MAX];int cnt,M;long long src[MAX],sum[MAX],f[MAX];int front,tail;inline double GetSlope(Point p1,Point p2){if(p1.x == p2.x)return INF;return (double)(p2.y - p1.y) / (p2.x - p1.x);}inline void Insert(long long x,long long y){Point temp(x,y);while(tail - front >= 2)if(GetSlope(q[tail],temp) < GetSlope(q[tail - 1],q[tail]))--tail;elsebreak;q[++tail] = temp;}inline Point GetAns(double slope){while(tail - front >= 2)if(GetSlope(q[front + 1],q[front + 2]) < slope)++front;elsebreak;return q[front + 1];}int main(){while(scanf("%d%d",&cnt,&M) != EOF) {for(int i = 1; i <= cnt; ++i) {scanf("%I64d",&src[i]);sum[i] = sum[i - 1] + src[i];}front = tail = 0;for(int i = 1; i <= cnt; ++i) {Insert(sum[i - 1],f[i - 1] + sum[i - 1] * sum[i - 1]);Point p = GetAns(sum[i] << 1);f[i] = p.y + sum[i] * sum[i] - (p.x * sum[i] << 1) + M;}printf("%I64d\n",f[cnt]);}return 0;}

0 0