“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-;
 const int MAXN = 1e6 + ;
 const int MAXM = 1e3 + ;
 const ll mod = ;

 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 = , r = tail - , ans = ;
     while(l <= r) {
         int mid = (l + r) >> ;
         ll up = dp[q[mid]] + sqr(sum[q[mid]]) - (dp[q[mid - ]] + sqr(sum[q[mid - ]]));
         ll down = 2.0 * (sum[q[mid]] - sum[q[mid - ]]);
         if(up <= down * x) l = mid + , ans = mid;
         else r = mid - ;
     }
     return q[ans];
 }

 int main() {
 #ifdef local
     freopen("data.txt", "r", stdin);
 #endif
     sum[] = dp[] = ;
     while(~scanf("%d",&n)) {
         for(int i = ; i <= n; i++) {
             scanf("%lld",&a[i]);
             sum[i] = sum[i - ] + a[i];
         }
         for(int i = ; i <= n; i++) scanf("%lld",&b[i]);
         tail = ;
         q[tail++] = ;
         for(int i = ; i <= n; i++) {
             int id = findd(1.0 * (sum[i] - b[i]));
             dp[i] = dp[id] + sqr(sum[i] - sum[id] - b[i]);
             while( < tail && getup(q[tail - ],q[tail - ]) * getdown(i,q[tail - ]) >= getup(i,q[tail - ]) * getdown(q[tail - ],q[tail - ]))
                 tail--;
             if(getdown(i,q[tail - ]) == ) k[tail] = 1e18;
             else k[tail] = 1.0 * getup(i,q[tail - ]) / getdown(i,q[tail - ]);
             q[tail++] = i;
         }
         printf("%lld\n",dp[n]);
     }
     return ;
 }