Problem Description Little D is really interested in the theorem of sets recently. There’s a problem that confused him a long time.  Let T be a set of integers. Let the MIN be the minimum integer in T and MAX be the maximum, then the cost of set T if…

#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<cmath> using namespace std; #define Maxn 1010 #define INF 0xfffffff *Maxn],sum[*Maxn]; *Maxn][*M…

T. E. Lawrence was a controversial figure during World War I. He was a British officer who served in the Arabian theater and led a group of Arab nationals in guerilla strikes against the Ottoman Empire. His primary targets were the railroads. A highl…

设d(i, j)为连通第i个点到第j个点的树的最小长度,则有状态转移方程: d(i, j) = min{ d(i, k) + d(k + 1, j) + p[k].y - p[j].y + p[k+1].x - p[i].x } 然后用四边形不等式优化之.. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> #d…

Division Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 999999/400000 K (Java/Others) Total Submission(s): 3984    Accepted Submission(s): 1527 Problem Description Little D is really interested in the theorem of sets recently. There’s a pro…

解题思路 第一步显然是将原数组排序嘛--然后分成一些不相交的子集,这样显然最小.重点是怎么分. 首先,我们写出一个最暴力的\(DP\): 我们令$F[ i ][ j ] $ 为到第\(i\)位,分成\(j\)组的代价,我们可以写出如下 $ DP$ for( LL i = 1; i <= N; ++i ) F[ i ][ 1 ] = sqr( A[ i ] - A[ 1 ] ); for( LL j = 2; j <= M; ++j ) for( LL i = j; i <= N; ++i…

环形石子合并问题. 有一种方法是取模,而如果空间允许的话(或者滚动数组),可以把长度为n个换拓展成长为2n-1的直线. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; + ; const int INF = 0x3f3f3f3f; int n; int a[maxn], sum[maxn]; int d[max…

题目链接:hdu 3480 Division 题意: 给你一个有n个数的集合S,现在让你选出m个子集合,使这m个子集合并起来为S,并且每个集合的(max-min)2 之和要最小. 题解: 运用贪心的思想,肯定首先将全部的数排好序,然后设dp[i][j]表示前j个数分为i个集合的最优解. 则有dp[i][j]=min{dp[i-1][k]+(a[j]-a[k+1])2}(0<k<j). 这样写出来是三层for的dp,考虑用斜率优化降维. 假设l<k<j,对于dp[i][j],k到j为…

HDU 2829 区间DP & 前缀和优化 & 四边形不等式优化 n个节点n-1条线性边,炸掉M条边也就是分为m+1个区间 问你各个区间的总策略值最少的炸法 就题目本身而言,中规中矩的区间DP问题 d p[i][j]表示前i个节点,分为j个区间的最优策略值 cost[i][j]为从i到j节点的策略值 所以dp[i][j] = min(dp[k-1][j-1] + cost[k][i] 但是复杂度太高了 可以优化的地方有: cost数组值得求取: 考虑到cost(i,j)=ΣAxAy (i≤…

题意:给定 n 个数,要你将其分成m + 1组,要求每组数必须是连续的而且要求得到的价值最小.一组数的价值定义为该组内任意两个数乘积之和,如果某组中仅有一个数,那么该组数的价值为0. 析:DP状态方程很容易想出来,dp[i][j] 表示前 j 个数分成 i 组.但是复杂度是三次方的,肯定会超时,就要对其进行优化. 有两种方式,一种是斜率对其进行优化,是一个很简单的斜率优化 dp[i][j] = min{dp[i-1][k] - w[k] + sum[k]*sum[k] - sum[k]*sum[…