这道题目是典型的DFS+记忆化搜索, DP思想.符合:含重叠子问题,无后效性等特点. #include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 105 int dp[MAXN][MAXN]; int a[MAXN][MAXN]; int n, k; int max(int a, int b) { return a>b ? a:b; } int dfs(int x, int y) { i…

这道题目是一道非常简单的贪心,但是我却修改了1h+.原因就是qsort的comp有bug.其实还是题目中的数据可以为0.除数为0真的要慎重啊.后来改为结构体,加一层循环选取最大值,果然ac啊.wa了几次,测试数据都过,还是wa.无语.这道题目一定要注意数据类型,double型. #include <stdio.h> #include <stdlib.h> #define MAXNUM 10050 typedef struct { int js, fs; double rate; }…

DP. #include <stdio.h> #include <string.h> #include <stdlib.h> #define MAXNUM 1005 typedef struct { int w, s; int index; } mouse_st; mouse_st mouses[MAXNUM]; int dp[MAXNUM], next[MAXNUM]; int comp(const void *a, const void *b) { mouse_st…

pid=1078">FatMouse and Cheese Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4811    Accepted Submission(s): 1945 Problem Description FatMouse has stored some cheese in a city. The city can…

HDU 1078 FatMouse and Cheese ( DP, DFS) 题目大意 给定一个 n * n 的矩阵, 矩阵的每个格子里都有一个值. 每次水平或垂直可以走 [1, k] 步, 从 (0, 0) 点开始, 下一步的值必须比现在的值大. 问所能得到的最大值. 解题思路 一般的题目只允许 向下 或者 向右 走, 而这个题允许走四个方向, 所以状态转移方程为 dp(x, y) = dp(nextX, nextY) + arr(x, y); dp 代表在 x, y 的最大值. 由于 下一…

FatMouse and Cheese Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8610    Accepted Submission(s): 3611 Problem Description FatMouse has stored some cheese in a city. The city can be considere…

FatMouse and Cheese FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 a…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1078 题意:每次仅仅能走 横着或竖着的 1~k 个格子.求最多能吃到的奶酪. 代码: #include <stdio.h> #include <ctime> #include <math.h> #include <limits.h> #include <complex> #include <string> #include <fun…

题意:给出n*n的二维矩阵,和k,老鼠每次最多走k步,问老鼠从起点(0,0)出发,能够得到的最大的数(即为将每走过一点的数都加起来的和最大)是多少 和上一题滑雪一样,搜索的方向再加一个循环 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<algorithm> using namespace std; typedef long long…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…