HDU 5492 Find a path 题意:给你一个矩阵求一个路径使得 最小. 思路: 方法一:数据特别小,直接枚举权值和(n + m - 1) * aver,更新答案. 方法二:用f[i][j][k]表示到达[i,j]是权值和为k时平方和的最大值,转移方程就是 f[i][j][k] = min(f[i][j][k], min(f[i - 1][j][k - a[i][j]] + sqr(a[i][j]), f[i][j - 1][k - a[i][j]] + sqr(a[i][j])));…

题意:在矩阵中,找一条路从 (1,1)->(n,m),使方差最小 思路: T = (N+M−1)∑N+M−1i=1(Ai−Aavg)2 将N + M - 1乘进去,即求1 ~ N+M-1,(N + M - 1)*A[i] - (A[i] + ..... + A[N]) 的和由于 假设Aavg可以是任何数,但只有当其是平均值时T才会最小(感觉别人都好厉害 /(ㄒoㄒ)/~~) #include <iostream> #include <cstdio> #include <…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5492 题目大意: 一个N*M的矩阵,一个人从(1,1)走到(N,M),每次只能向下或向右走.求(N+M-1)ΣN+M-1(Ai-Aavg)2最小.Aavg为平均值. (N,M<=30,矩阵里的元素0<=C<=30) 题目思路: [动态规划] 首先化简式子,得原式=(N+M-1)ΣN+M-1(Ai2)-(ΣN+M-1Ai)2 f[i][j][k]表示走到A[i][j]格子上,此时前i+j-1…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5492 题目大意是有一个矩阵,从左上角走到右下角,每次能向右或者向下,把经过的数字记下来,找出一条路径是这些数的方差最小. Find a path Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 961    Accepted Submissi…

Beautiful Now Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1876    Accepted Submission(s): 707 Problem Description Anton has a positive integer n, however, it quite looks like a mess, so he…

Find a path Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 549264-bit integer IO format: %I64d      Java class name: Main Frog fell into a maze. This maze is a rectangle containing N rows and M columns. Each…

Fxx and string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 507    Accepted Submission(s): 224 Problem Description Young theoretical computer scientist Fxx get a string which contains lowerc…

思路:用map将每个字符串与一个数字进行对应,然后暴力统计就好了 #include<cstring> #include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<map> #include<vector> #include<string> #define Maxn 2010 using namespace st…

http://acm.hdu.edu.cn/showproblem.php?pid=4952 给定x,k,i从1到k,每次a[i]要是i的倍数,并且a[i]大于等于a[i-1],x为a0 递推到下一个a==之前的a即可跳出循环 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <queue&g…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4708 思路:由于N不大,并且我们可以发现通过旋转得到的4个对角线的点的位置关系,以及所要旋转的最小步数,然后把所有的符合条件的都加入数组中,最后排序即得. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct No…