【斜率优化】HDU 3507 Print Article

通道

题意：给定一个长度为n的序列，和一个常数m，我们可以将序列分成随意段，每段的权值为sum(arr[i]) + C(l<=i<=r)，求一种划分方法使得整个序列的权值最小

思路：

方程：dp[i] = min(dp[j] + (sum[i]-sum[j])^2 + m) (j < i);
     则有：dp[i] = dp[j] + (sum[i]-sum[j])^2 + m = dp[j] + sum[i] * sum[i] + sum[j] * sum[j] - 2 * sum[i] * sum[j] + m;
     设y =  dp[j] + sum[j] * sum[j],x = sum[j]，那么原式等于:dp[i] = y + 2 * sum[i] * x + m + sum[i] * sum[i]

代码：

#include <cstdio>
#include <cstring>

typedef long long ll;

const int MAX_N = 500007;
 
struct Point {
    ll x, y, c;
    Point () {
        
    }
    Point (ll _x, ll _y, ll _c) {
        x = _x;
        y = _y;
        c = _c;
    }
};

int n, m;
int a[MAX_N];
Point p[MAX_N];
ll sum[MAX_N], dp[MAX_N];
int que[MAX_N], head, tail;
 
int Check(int x, int y, int z) { 
    Point p0 = p[x], p1 = p[y], p2 = p[z]; 
    return (p0.x -p1.x) * (p0.y - p2.y) <= (p0.y - p1.y) * (p0.x - p2.x);
} 
int NotBest(int x, int y, int k) { 
    Point p0 = p[x], p1 = p[y]; 
    return p0.y - k * p0.x > p1.y - k * p1.x; 
} 

ll Posi() { 
    head = tail = 0; 
    que[tail] = 0; 
    p[0].x = p[0].y = 0; 
    for (int i = 1; i <= n; ++i) { 
        p[i].x = sum[i - 1]; 
        p[i].y = dp[i-1] + sum[i - 1] * sum[i - 1]; 
        while (head <= tail - 1 &&  Check(que[tail - 1], que[tail], i)) tail--; 
        que[++tail] = i; 
        while (head + 1 <= tail && NotBest(que[head], que[head + 1], 2 * sum[i])) head++; 
        int k = que[head]; 
        dp[i] = p[k].y - 2 * sum[i] * p[k].x + sum[i] * sum[i] + m; 
    } 
    return dp[n]; 
} 

ll Nega() { 
    ll Min = 0;
    for (int i = 1; i <= n; ++i) {
        Min = -1;
        for (int j = que[i - 1]; j < i; ++j) {
            if (Min == -1 || dp[j] + (sum[i] - sum[j]) * (sum[i] - sum[j]) < Min) {
                Min = dp[j] + (sum[i] - sum[j]) * (sum[i] - sum[j]); 
                que[i] = j;
            }
        }
        dp[i] = Min + m;
    }
    return dp[n]; 
} 
 
 
int main() { 
    while (2 == scanf("%d%d", &n, &m)) { 
        for (int i = 1; i <= n; ++i) 
            scanf("%d",&a[i]), sum[i] = a[i] + sum[i - 1]; 
        ll ans = Nega(); 
        printf("%lld\n", ans); 
    } 
    return 0;
}