POJ 3176 Cow Bowling 题目简化即为从一个三角形数列的顶端沿对角线走到底端,所取得的和最大值 7 * 3 8 * 8 1 0 * 2 7 4 4 * 4 5 2 6 5 该走法即为最大值 分析:简单的动态规划,从上往下一层一层的考虑,对于每一行的最左边和最右边只有一种走法,只需要简单的相加, 对于中间的数要考虑是加上左上角的数还是加右上角的数,加上两者中的较大者 代码: #include<iostream> #include<cstdio> #include<…

id=1163">链接:poj 1163 题意:输入一个n层的三角形.第i层有i个数,求从第1层到第n层的全部路线中.权值之和最大的路线. 规定:第i层的某个数仅仅能连线走到第i+1层中与它位置相邻的两个数中的一个. 状态方程:f[i][j]=max(f[i-1][j-1],f[i-1][j])+a[i][j]; 1163代码: #include<stdio.h> #include<string.h> int a[105][105],f[105][105]; int…

题目链接:http://poj.org/problem?id=3176 思路分析:基本的DP题目:将每个节点视为一个状态,记为B[i][j], 状态转移方程为 B[i][j] = A[i][j] + Max( B[i+1][j], B[i+1][j+1] ); 代码如下: #include <stdio.h> + ; int A[MAX_N][MAX_N], B[MAX_N][MAX_N]; int Max( int a, int b ) { return a > b ? a : b;…

Cow Bowling Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13016   Accepted: 8598 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard…

Cow Bowling Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16448 Accepted: 10957 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bow…

题 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this: 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 Then the other cows travers…

北大教你怎么打保龄球 题目很简单的,我就不翻译了,简单来说就是储存每一行的总数,类似于状态压缩 #include <stdio.h> #include <stdlib.h> #define MAX(a,b) a>b?a:b ][]; void Search(const int); int main(void) { int N, i, line; while (~scanf("%d", &N)) { Cow_Map[][] = ; ; line &l…

题意:给定一个金字塔,第 i 行有 i 个数,从最上面走下来,只能相邻的层数,问你最大的和. 析:真是水题,学过DP的都会,就不说了. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <…

Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this: Then the other cows traverse the triangle starting from its…

Cow Bowling Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13464   Accepted: 8897 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard…

题 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this: 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 Then the other cows travers…

题意:给一个三角形形状的数字,从上到下,要求数字和最大 思路 :dp dp[i+1][j]=max(dp[i+1][j],dp[i][j]+score[i+1][j]) dp[i+1][j+1]=max(dp[i+1][j],dp[i][j]+score[i+1][j+1] 在最后一行进行比较,找到最大值输出 对上面思路的解释: 对于每个位置都是由上面一个位置加当前位置的最大值组成,所以有了上面的递推公式 解决问题的代码: #include <iostream> #include <cs…

牛保龄球 直接中文了 Descriptions 奶牛打保龄球时不使用实际的保龄球.它们各自取一个数字(在0..99范围内),然后排成一个标准的保龄球状三角形,如下所示: 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 然后其他奶牛从其尖端开始穿过三角形并“向下”移动到两个对角相邻的奶牛中的一个,直到到达“底部”行.奶牛的得分是沿途参观的奶牛数量的总和.得分最高的母牛赢得了那个框架. 给定具有N(1 <= N <= 350)行的三角形,确定可实现的最高可能总和. input 第1行:…

动态规划:多阶段决策问题,每步求解的问题是后面阶段问题求解的子问题,每步决策将依赖于以前步骤的决策结果.(可以用于组合优化问题) 优化原则:一个最优决策序列的任何子序列本身一定是相当于子序列初始和结束状态的最优决策序列. 只有满足优化原则的问题才可以利用动态算法进行求解,因为只有全局最优解法等于其每个子问题的最优才可以分阶段进行求解. The cows don't use actual bowling balls when they go bowling. They each take a nu…

POJ 3176 Cow Bowling 链接: http://poj.org/problem?id=3176 这道题可以算是dp入门吧.可以用一个二维数组从下向上来搜索从而得到最大值. 优化之后可以直接用一维数组来存.(PS 用一维的时候要好好想想具体应该怎么存,还是有技巧的) #include<iostream> #include<cstring> #include<cmath> #include<cstdio> using namespace std;…

OJ上的一些水题(可用来练手和增加自信) (POJ 3299,POJ 2159,POJ 2739,POJ 1083,POJ 2262,POJ 1503,POJ 3006,POJ 2255,POJ 3094) 初期: 一.基本算法: 枚举. (POJ 1753,POJ 2965) 贪心(POJ 1328,POJ 2109,POJ 2586) 递归和分治法. 递推. 构造法.(POJ 3295) 模拟法.(POJ 1068,POJ 2632,POJ 1573,POJ 2993,POJ 2996) 二…

著名题单,最初来源不详.直接来源:http://blog.csdn.net/a1dark/article/details/11714009 OJ上的一些水题(可用来练手和增加自信) (POJ 3299,POJ 2159,POJ 2739,POJ 1083,POJ 2262,POJ 1503,POJ 3006,POJ 2255,POJ 3094) 初期: 一.基本算法: 枚举. (POJ 1753,POJ 2965) 贪心(POJ 1328,POJ 2109,POJ 2586) 递归和分治法. 递…

POJ 1852 Ants POJ 2386 Lake Counting POJ 1979 Red and Black AOJ 0118 Property Distribution AOJ 0333 Ball POJ 3009 Curling 2.0 AOJ 0558 Cheese POJ 3669 Meteor Shower AOJ 0121 Seven Puzzle POJ 2718 Smallest Difference POJ 3187 Backward Digit Sums POJ 3…

先来看一下经典的背包问题吧 http://www.cnblogs.com/Kalix/p/7617856.html  01背包问题 https://www.cnblogs.com/Kalix/p/7622102.html 完全背包问题 https://blog.csdn.net/mystery_guest/article/details/51878140      多重背包二进制优化 1.https://cn.vjudge.net/problem/12304/origin    POJ 3176…

The Triangle Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 41169   Accepted: 24882 Description 7 3 8 8 1 0 2 7 4 4 4 5 2 6 5 (Figure 1) Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed…

终于写完了POJ的DP专题,然而都是水题233 这次也把题目分了一下,先挑3道特别简单的讲一下 2533 题意:求最长上升子序列. 很简单,用一般的DP或者二分优化都可以过去 这里懒得写一般DP了,其实就是用f[i]表示前i个数中LIS的数量,那么在i之前找一个j,满足a[j]<a[j]并且f[j]最大,然后转移即可 边界条件:f[i]=1 二分的算法我也有所提及 二分CODE #include<cstdio> using namespace std; const int N=1005;…

Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7644   Accepted: 2798   Special Judge Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets…

Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   Special Judge Description The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000…

The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22286   Accepted: 8603   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…

Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37427   Accepted: 16288 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the…

Corn Fields Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9806   Accepted: 5185 Description Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yumm…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050   Accepted: 10989 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11939   Accepted: 7493 Description Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital le…

Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17898   Accepted: 9197 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names t…

题目: poj 2352 Stars 数星星 题意:已知n个星星的坐标.每个星星都有一个等级,数值等于坐标系内纵坐标和横坐标皆不大于它的星星的个数.星星的坐标按照纵坐标从小到大的顺序给出,纵坐标相同时则按照横坐标从小到大输出. (0 <= x, y <= 32000) 要求输出等级0到n-1之间各等级的星星个数. 分析: 这道题不难想到n平方的算法,即从纵坐标最小的开始搜,每次找它前面横坐标的值比它小的点的个数,两个for循环搞定,但是会超时. 所以需要用一些数据结构去优化,主要是优化找 横坐…