hdu3507Print Article (斜率dp水题)_hdu 3507 kuangbin-优快云博客

本文链接：https://blog.youkuaiyun.com/guognib/article/details/13658671

本文探讨了动态规划中的优化技巧，特别是通过利用单调性原理来提高算法效率。介绍了如何在信息学竞赛中巧妙地结合数形概念，实现算法性能的提升。

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论文：NOI2004年周源的论文《浅谈数形结合思想在信息学竞赛中的应用》，

JSOI2009集训队论文《用单调性优化动态规划》

参考：http://www.cnblogs.com/kuangbin/archive/2012/08/26/2657650.html

//#pragma warning (disable: 4786)
//#pragma comment (linker, "/STACK:16777216")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)

typedef long long LL;
const int INF = 1000000007;
const double eps = 1e-10;
const int MAXN = 500010;

int n, M;
int dp[MAXN];
int sum[MAXN];
int q[MAXN];
int be, ed;

int Up(int k, int j)
{
    return dp[j] + sum[j] * sum[j] - (dp[k] + sum[k] * sum[k]);
}
int Down(int k, int j)
{
    return (sum[j] - sum[k]) * 2;
}

int getDp(int i, int j)
{
    return dp[j] + (sum[i] - sum[j]) * (sum[i] - sum[j]) + M;
}

int solve()
{
    dp[0] = 0;
    be = 1;
    ed = 0;
    q[++ed] = 0;///

    for (int i = 1; i <= n; i++)
    {
        while (be < ed && Up(q[be], q[be + 1]) <= sum[i] * Down(q[be], q[be + 1]))
            be++;
        int j = q[be];
        dp[i] = getDp(i, j);
        while (be < ed && Up(q[ed - 1], q[ed]) * Down(q[ed], i) >= Up(q[ed], i) * Down(q[ed - 1], q[ed]))
            ed--;
        q[++ed] = i;
    }
    return dp[n];
}

int main ()
{
    int x;
    while (~RII(n, M))
    {
        sum[0] = 0;
        FE(i, 1, n)
        {
            RI(x);
            sum[i] = sum[i - 1] + x;
        }
        printf("%d\n", solve());
    }
    return 0;
}