“景驰科技杯”2018年华南理工大学程序设计竞赛 G. Youhane as "Bang Riot"（斜率DP）

题目链接：https://www.nowcoder.com/acm/contest/94/G

题意：中文题目，见链接

题解：设 sum[i] 为 a[i] 的前缀和，可得公式

dp[i] = min( dp[j] + ( sum[i] - sum[j] -  b[i] ) ^ 2 )

          = min( dp[j] + sum[j] ^ 2 + 2 * ( sum[i] - b[i] ) * ( sum[i] - sum[j] ) + sum[i] - b[i] ) ^ 2 

设 y = dp[j] + sum[j] ^ 2，k = sum[i] - b[i]，x = sum[i] - sum[j]，b = sum[i] - b[i] ) ^ 2

有 y = 2 * k * x + b

因为 k 的正负不确定，但 y 和 x 符合单调性，故可以二分找出第一个比当前点斜率大的点来维护下凸性（注意最好手写二分，使用lower_bound可能会出错）

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define mst(a,b) memset((a),(b),sizeof(a))
#define mp(a,b) make_pair(a,b)
#define pi acos(-1)
#define pii pair<int,int>
#define pb push_back
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 1e6 + 10;
const int MAXM = 1e3 + 10;
const ll mod = 100000073;
 
int n,tail;
int q[MAXN];
ll a[MAXN],b[MAXN],sum[MAXN],dp[MAXN];
double k[MAXN];
 
ll sqr(ll x) {
    return x * x;
}
 
ll getup(int j,int k) {
    return dp[j] + sqr(sum[j]) - (dp[k] + sqr(sum[k]));
}
 
ll getdown(int j,int k) {
    return 2ll * (sum[j] - sum[k]);
}
 
int findd(double x) {
    int l = 0, r = tail - 1, ans = 0;
    while(l <= r) {
        int mid = (l + r) >> 1;
        ll up = dp[q[mid]] + sqr(sum[q[mid]]) - (dp[q[mid - 1]] + sqr(sum[q[mid - 1]]));
        ll down = 2.0 * (sum[q[mid]] - sum[q[mid - 1]]);
        if(up <= down * x) l = mid + 1, ans = mid;
        else r = mid - 1;
    }
    return q[ans];
}
 
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
#endif
    sum[0] = dp[0] = 0;
    while(~scanf("%d",&n)) {
        for(int i = 1; i <= n; i++) {
            scanf("%lld",&a[i]);
            sum[i] = sum[i - 1] + a[i];
        }
        for(int i = 1; i <= n; i++) scanf("%lld",&b[i]);
        tail = 0;
        q[tail++] = 0;
        for(int i = 1; i <= n; i++) {
            int id = findd(1.0 * (sum[i] - b[i]));
            dp[i] = dp[id] + sqr(sum[i] - sum[id] - b[i]);
            while(1 < tail && getup(q[tail - 1],q[tail - 2]) * getdown(i,q[tail - 1]) >= getup(i,q[tail - 1]) * getdown(q[tail - 1],q[tail - 2]))
                tail--;
            if(getdown(i,q[tail - 1]) == 0) k[tail] = 1e18;
            else k[tail] = 1.0 * getup(i,q[tail - 1]) / getdown(i,q[tail - 1]);
            q[tail++] = i;
        }
        printf("%lld
",dp[n]);
    }
    return 0;
}