玩具装箱

http://www.lydsy.com/JudgeOnline/problem.php?id=1010

斜率优化DP

设dp[i]表示前i个玩具装箱所需的最小耗费

dp[i] = min(dp[j]+(i-j-1+∑C[k]-L)^2)

设sc[i] = ∑C[k] ( 1<= k <= i)

则有

   dp[j]+(i-j-1+∑C[k]-L)^2

= dp[j]+(i-j-1+sc[i]-sc[j]-L)^2

= dp[j]+((i-1+sc[i]-L)-(sc[j]+j))^2

= dp[j]+(i-1+sc[i]-L)^2+(sc[j]+j)^2-2*(i-1+sc[i]-L)*(sc[j]+j)

用y表示dp[j]+(sc[j]+j)^2

用x表示sc[j]+j

用a表示2*(i-a+sc[i]-L)

G = -ax+y,其中a是单调递增的，满足斜率优化的条件

则dp[i] = min(G)+i-1+sc[i]-L    

  #include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <cstring>#include <stack>#include <cctype>#include <utility>   #include <map>#include <string>  #include <climits> #include <set>#include <string>    #include <sstream>#include <utility>   #include <ctime>#include <bitset> using std::priority_queue;using std::vector;using std::swap;using std::stack;using std::sort;using std::max;using std::min;using std::pair;using std::map;using std::string;using std::cin;using std::cout;using std::set;using std::queue;using std::string;using std::stringstream;using std::make_pair;using std::getline;using std::greater;using std::endl;using std::multimap;using std::deque;using std::unique;using std::lower_bound;using std::random_shuffle;using std::bitset;using std::upper_bound;using std::multiset; typedef long long LL;typedef unsigned long long ULL;typedef pair<int, int> PAIR;typedef multimap<int, int> MMAP;typedef LL TY;typedef long double LF; const int MAXN(50010);const int MAXM(100010);const int MAXE(100010);const int MAXK(6);const int HSIZE(31313);const int SIGMA_SIZE(26);const int MAXH(19);const int INFI((INT_MAX-1) >> 1);const ULL BASE(31);const LL LIM(10000000);const int INV(-10000);const int MOD(20100403);const double EPS(1e-7);const LF PI(acos(-1.0)); template<typename T> void checkmax(T &a, T b){if(b > a) a = b;}template<typename T> void checkmin(T &a, T b){if(b < a) a = b;}template<typename T> T ABS(const T &a){return a < 0? -a: a;} int que[MAXN];int front, back;LL sc[MAXN]; LL table[MAXN];LL Y(int ind){return table[ind]+(ind+sc[ind])*(ind+sc[ind]);}LL X(int ind){return ind+sc[ind];} int main(){    int N, L;    while(~scanf("%d%d", &N, &L))    {        for(int i = 1; i <= N; ++i)        {            scanf("%lld", sc+i);            sc[i] += sc[i-1];        }        LL ans = 1e12;        front = 0;        back = -1;        que[++back] = 0;        for(int i = 1; i <= N; ++i)        {            LL temp =i-1+sc[i]-L;             LL a = 2*temp;            while(back-front > 0 && (-a)*X(que[front+1])+Y(que[front+1]) <= (-a)*X(que[front])+Y(que[front])) ++front;            table[i] = (-a)*X(que[front])+Y(que[front])+temp*temp;            while(back-front > 0 && (Y(i)-Y(que[back-1]))*(X(i)-X(que[back])) >= (Y(i)-Y(que[back]))*(X(i)-X(que[back-1]))) --back;            que[++back] = i;        }        printf("%lld\n", table[N]);    }    return 0;}