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<…

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 <…

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…

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…

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…

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;…

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…

题意:给你两个数,求所有的数位的积的和. 析:太水了,没的说,可以先输入边算,也可以最后再算,一样.. 代码如下: #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #inclu…

POJ 1488 题目大意:给定一篇文章,将它的左引号转成 ``(1的左边),右引号转成 ''(两个 ' ) 解题思路:水题,设置一个bool变量标记是左引号还是右引号即可 /* POJ 1488 Tex Quotes --- 水题 */ #include <cstdio> #include <cstring> int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #…

A.Beru-taxi 水题:有一个人站在(sx,sy)的位置,有n辆出租车,正向这个人匀速赶来,每个出租车的位置是(xi, yi) 速度是 Vi;求人最少需要等的时间: 单间循环即可: #include<iostream> #include<algorithm> #include<string.h> #include<stdio.h> #include<math.h> #include<vector> using namespace…

http://poj.org/problem?id=3461 直接KMP就好.水题 #include<cstdio> #include<cstring> const int MAXN=10000+10; const int MAXM=1000000+10; char P[MAXN],T[MAXM]; int f[MAXN],n,m,ans; void getFail() { f[0]=f[1]=0; for(int i=1;i<n;i++){ int j = f[i]; wh…

Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 110991   Accepted: 34541 Description Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical,…

487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 236746   Accepted: 41288 Description Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phras…

Hangover Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 99450   Accepted: 48213 Description How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're as…

一.Description Suppose you are reading byte streams from any device, representing IP addresses. Your task is to convert a 32 characters long sequence of '1s' and '0s' (bits) to a dotted decimal format. A dotted decimal format for an IP address is form…

一.Description The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days (the second and third days of service), the knight receives two gold coins. On each of the ne…

题目 http://poj.org/problem?id=1837 题意 单组数据,有一根杠杆,有R个钩子,其位置hi为整数且属于[-15,15],有C个重物,其质量wi为整数且属于[1,25],重物与重物之间,钩子与钩子之间彼此不同.忽略杠杆及重心的影响,有多少种方式使得全部重物都挂上钩子(某些钩子可能挂若干个重物)后杠杆平衡? 思路 由于状态比较小,即使n的五次方也足以承受,而且任意时刻杠杆的状态在[-15 * 25 * 20, 15 * 25 * 20]之间,所以可以直接穷举状态. 感想…

<题目链接> 题目大意: 给定一颗树,求出树的直径. 解题分析:树的直径模板题,以下程序分别用树形DP和两次BFS来求解. 树形DP: #include <cstdio> #include <algorithm> using namespace std; ; struct Edge{ int to,val,nxt; Edge(,,):to(_to),val(_val),nxt(_nxt){} }e[N<<]; int n,m,cnt,ans; int dp1…

题目:http://poj.org/problem?id=1035 还是暴搜 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cmath> #include<map> #include&…

题意:给定一个完全由小写字母组成的字符串s,对每个字母比如x(或a,b,c...z),在字符串中添加或者删除它分别需要花费c1['x']和c2['x']的代价,问将给定字符串变成回文串所需要的最少代价为多少. 解法:设d[i][j]表示将字符串中从第i位至第j位变成回文串所需要的代价.若s[i] == s[j],d[i][j] = d[i+1][j-1]:否则的话,有四种处理方法. 对xa.......by,可以将其变为xa......b,yxa.....by,a.......by,xa....…

再思考一下好的方法,水过,数据太弱! 本来不想传的! #include <iostream> using namespace std; #define MAX 702 /*284K 422MS*/ typedef struct _point { int x; int y; }point; point p[MAX]; bool judge(point a,point b,point c) { return (a.y-b.y)*(c.x-b.x)-(c.y-b.y)*(a.x-b.x); } in…

题目 http://poj.org/problem?id=2002 题意 已知平面内有1000个点,所有点的坐标量级小于20000,求这些点能组成多少个不同的正方形. 思路 如图,将坐标按照升序排列后,首先枚举p1,p2, 并判断p2是否在p1正下方或者左上角(因为每个正方形只有一条最右边或者是右下的边),按照下图计算p3,p4,判断p3,p4是否存在即可. 感想 排序时要注意和左上角这个信息相符,刚写完时用的是左下角,与升序排序不符合,会遗失部分正方形. 代码 #include <cstdio…

题意:给定一棵树,然后让你找出它的直径,也就是两点中的最远距离. 析:很明显这是一个树上DP,应该有三种方式,分别是两次DFS,两次BFS,和一次DFS,我只写了后两种. 代码如下: 两次BFS: #include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <queue> using namespace std; const int max…

题目链接:POJ 2365 Rope Rope Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7488   Accepted: 2624 Description Plotters have barberically hammered N nails into an innocent plane shape, so that one can see now only heads. Moreover, pursuing th…

一.Description Bill and Ted are taking a road trip. But the odometer in their car is broken, so they don't know how many miles they have driven. Fortunately, Bill has a working stopwatch, so they can record their speed and the total time they have dri…

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…

基本上还是01背包,首先注意必须正好花光钱,所以初始化时除了dp[0]以外其他都要设置成inf,然后因为求方案数,所以基本方程为dp[i] = dp[i-x] + dp[i],再根据inf进行一些特殊处理即得解 #include <cstdio> #include <cstring> #include <algorithm> + ; ; int dp[maxm]; int n, m; int x; int main () { scanf("%d %d"…

最基本的01背包,不需要太多解释,刚学dp的同学可以参见dd大牛的背包九讲,直接度娘“背包九讲”即可搜到 #include <cstdio> #include <cstring> #include <algorithm> + ; int dp[maxn]; int t, m; int s1, v1; int main () { scanf("%d %d", &t, &m); ; i <= m; i++) { scanf(&quo…

题意:给定一个n*m的矩阵,让你判断有多少个连通块. 析:用DFS搜一下即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring&…

关于鸽笼原理的知识看我写的另一篇博客 http://blog.csdn.net/u011026968/article/details/11564841 (需要说明的是,我写的代码在有答案时就输出结果了,但OJ也是从文件读入,所以乍一看我的好像在没输入完就有结果了,但OJ不知道,其实我是直接拿poj3370的代码AC的,32MS,O(∩_∩)O) 直接贴代码 #include<cstdio> #include<cstring> using namespace std; #define…