Two Sawmills(锯木厂选址)

http://main.edu.pl/en/archive/ceoi/2004/two

斜率优化DP,应该说是第一道斜率优化DP了,推公式的时候各种坑,还是参照了hzq神牛的思路，细节方面稍有不同,

为了思维方便，我先将给出的序列翻转了,也就是把从山顶到山下的点顺序边成了从山下到山顶,编号从1开始,第一个点即为海拔最低的伐木场,所以共有n+1个点,w[i]表示第i个点的重量,dist[i]表示第i个点到第一个点的距离,dp[i]表示把第二个伐木场建到第i个点的最优解

则有 

         dp[i] = min(S[1, j-1]+S[j, i-1]+S[i, n+1])    1

其中  S[l, r] = ∑w[i]*(dist[i]-dist[i])  (l <= i <= r)  2 

再设sw[i] = ∑ w[j] (1 <= j <= i),                         3

        swd[i] = ∑ w[j]*dist[j] (1 <= j <= i)             4

则有 S[l, r] = swd[r]-swd[l-1]-(sw[r]-sw[l-1])*dist[l];

带入1式后化简可得

dp[i] = min(swd[n+1]-swd[0]-(sw[n+1]-sw[i-1])*dist[i]-(sw[i-1]-sw[j-1])*dist[j]-(sw[j-1]-sw[0])*dist[1])

由于dist[1]等于0,所以

dp[i] = min(-(sw[i-1]-sw[j-1])*dist[j])+swd[n+1]-swd[0]-(sw[n+1]-sw[i-1])*dist[i]

以为后面的部分是与j无关的常量，所以不会影响决策,可以忽略掉

所以要求解只剩min((-sw[i-1])*dist[j]+sw[j-1]*dist[j])了

由于-sw[i-1]在第i轮决策中可以看成常数记为a,用x表示dist[j],y表示sw[j-1]*dist[j],

则优化目标G = min(-ax+y),设x*,y*为最优解(注意最优解实际上只有j决定)

则有G = -ax*+y*,移项后可得y* = ax*+G,由于a > 0所以，这相当于一条斜率已知的直线向上移动直到与之前任意一个点(x, y) 想切,此时的到的G是最优解,不难发现最优决策点构成一条下凸线,而且由于a单调递增,而下凸线的斜率也是单调递增的，所以可以舍弃当前最优决策之前的所有点，因为这些点都不可能成为之后决策的最优点了。

#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(20010);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;}LL X[MAXN], Y[MAXN];int que[MAXN];int front, back;LL dist[MAXN], sw[MAXN];int main(){int n;while(~scanf("%d", &n)){++n;  //一共n+1个点,下标1从开始LL swd = 0;for(int i = n; i >= 2; --i)scanf("%I64d%I64d", sw+i, dist+i);for(int i = 1; i <= n; ++i){dist[i] += dist[i-1];swd += dist[i]*sw[i];sw[i] += sw[i-1];}for(int i = 1; i <= n; ++i){X[i] = dist[i];Y[i] = sw[i-1]*dist[i];}LL ans = 1e11;front = 0;back = -1;que[++back] = 1;que[++back] = 2; for(int i = 3; i <= n; ++i){while(back-front > 0 && (-sw[i-1])*X[que[front+1]]+Y[que[front+1]] <= (-sw[i-1])*(X[que[front]])+Y[que[front]]) ++front;checkmin(ans, swd-(sw[n]-sw[i-1])*dist[i]+((-sw[i-1])*(X[que[front]])+Y[que[front]]));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("%I64d\n", ans);}return 0;}